Geometric Progressions

(i) Let a = 4, b = -2, c = 1.

In GP, b²=ac

⇒ (-2)² = 4.1

⇒ 4 = 4

Common ratio = r

(ii) Let , b = -6, c = -54.

In GP, b²=ac

⇒

⇒ 36 = 36

Common ratio = r

(iii) Let a = a,

In GP, b²=ac

⇒

⇒

Common ratio

(iv) Let

In GP, b²=ac

⇒

⇒

Common ratio

Put n = 1,2,3,4…

a₁, a₂, a₃, a₄

⇒

If a₁, a₂, a₃ …

⇒ (a₂)² = a₁.a₃

⇒

⇒

So, It is GP.

Common ratio

T_n = ar^n-1

a = 4,

∴ T₉ = 4.(4^9-1)

= 4.4⁸

= 4⁹

∴ The 9 term is 4⁹

T_n = ar^n-1

∴ T₉

∴ The 10 term is .

T_n = ar^n-1

∴ T₈

∴ The 10 term is .

T_n = ar^n-1

∴ T₁₂

= a⁴¹x⁴¹

∴ The 12 term is a⁴¹x⁴¹.

T_n = ar^n-1

∴ T_n =

∴ The n term is .

T_n = ar^n-1

∴ T₁₀

∴ The 10 term is .

Nth term from the end is given by,

N

Where l = last term, n = nth term, and r = common ratio.

L = last term = 162

common ratio

n = 4

∴ N

⇒ N

⇒ 6

∴ 4^th term from last is 6.

T_n = ar^n-1

, T_n = 12.5, n=?

∴

⇒

⇒

⇒

⇒ n = 6

T_n = ar^n-1

∴

⇒

⇒

⇒

⇒ n = 11

T_n = ar^n-1

∴

⇒ 64√2 = (√2)ⁿ

⇒ n = 13

T_n = ar^n-1

∴

729 = (√3)ⁿ

⇒ n = 12

T_n = ar^n-1

∴

⇒

⇒ n = 9

T_n = ar^n-1

⇒

⇒

⇒ n = 9

Nth term from the end is given by

N =

Where, l = last term, n = nth term, and r = common ratio.

L = last term

common ratio

n = 4

∴ N =

⇒ N =

⇒

∴ 4^th term from last is .

T_n = ar^n-1

a =a, r =?, T_n = 27 n=4

a =a, r =?, T_n = 729 n=7

∴ 27 = a.r^4-1

⇒ 27 =a.r³…(1)

∴ 729 = a.r^7-1

⇒ 729 = a.r⁶…(2)

Divide (2) by (1) we get

⇒

⇒ r³ = 27

⇒ r = 3

Substituting r in 1 we get

a = 1

∴ GP = 1,3,9…

T_n = ar^n-1

In the GP, the 7th term is 8 times the 4th term. So

ar⁶ = 8ar³, or

r³ = 8 or r = 2.

ar⁴ = 48,

a.16 = 48

a = 3

The first term is 3 and r = 2

∴ GP = 3, 6, 12, …

GP is given by

a, ar, ar², ⋯, ar^n-1

In the sequence 5, 10, 20, 40, …

First term, a = 5

Common ratio, r

Equate the term to be found with the nnth term.

ar^n-1=1280

5 × 2^n-1 =1280

2^n-1 = 256

n - 1 = 8

n =9

∴ 9^th term is equal.

T_n = ar^n-1

According to the question,

⇒ T₅ = a.r⁴

⇒ T₈ = a.r⁷

⇒ T₁₁ = a.r¹⁰

To Prove: q²=p.s

⇒ T₅ = a.r⁴ = p

⇒ T₈ = a.r⁷ = q

⇒ T₁₁ = a.r¹⁰=s

q²= (a.r⁷)² = a².r¹⁴…(1)

p.s = (a.r⁴)(a.r¹⁰) = a².r¹⁴…(2)

∴ from (1) and (2) we get

⇒ q²=p.s

Hence, Proved.

a = -1

T₄ = (T₂)²

T_n = ar^n-1

∴ a.r³ = (ar)²

r³ = a.r²

⇒ r = a

∴ r =-1

T₇ = ar^7-1

= (-1).r⁶

= -1.(-1)⁶

= -1.

T_n = ar^n-1

a = a, r =?, T_n = 24 n = 3

a = a, r =?, T_n = 192 n = 6

∴ 24 = a.r^3-1

⇒ 24 = a.r²…(1)

∴ 192 = a.r^6-1

⇒ 192 = a.r⁵…(2)

Divide (2) by (1) we get

⇒

⇒ r³ = 8

⇒ r = 2

Substituting r in 2 we get

a = 6

T₁₀ = 6.2^10-1

= 6.2⁹

= 3072.

We observe that the left side of the inequality could be written like:

(ap - b)² + (bp - c)² + (cp - d)² ≥ 0 as each of these 3 terms is a

perfect square..........(1)

But by the given condition:

(ap - b)² + (bp - c)² + (cp - d)² ≤ 0 ........(2)

Therefore the conditions (1) and(2) can be satisfying iff the sum, (ap - b)² + (bp - c)² + (cp - d)² = 0 which is possible iff each of the terms,

(ap - b) = (bp - c) = (cp - d) is equal to zero. So,

ap - b = 0, Or a/b = k.

bp - c = 0. Or b/c = k.

cp - d = o, Or a/d =k.

Therefore, a/ b = b/c = c/d = k the common ratio of the terms a, b, c, and d.

So a,b,c,and d are in geometric progression .

Given:

To Prove: a, b, c, and d are in G.P

Proof:

Applying componendo and dividend to the given expression, we get,

Clearly, a, b, c, and d are in G.P.

Hence, Proved.

Given: p^th and q^th terms of a G.P. are q and p

Formula Used: T_n = ar^n-1

So, we get,

q = ar^p-1 …(1)

p = ar^q-1…(2)

To Prove:

Proof:

Divide (1) by (2), we get

⇒

⇒

Substituting we get,

⇒

L.H.S = ar^p+q-1

L.H.S = R.H.S

Hence, Proved.

Let the three numbers be .

∴ According to the question

⇒ …(1)

⇒ …(2)

From 2 we get,

⇒ a³ = 3375

∴ a = 15.

From 1 we get,

⇒

⇒ a + ar + ar² = 65r …(3)

Substituting a = 15 in 3 we get

⇒ 15 + 15r + 15r² = 65r

⇒ 15r² – 50r + 15 = 0…(4)

Dividing (4) by 5 we get

⇒ 3r² – 10r + 3 = 0

⇒ 3r² – 9r – r + 3 = 0

⇒ 3r(r – 3) – 1(r – 3) = 0

∴ r = 3 or r = 1/3

∴ Now the equation will be

⇒

⇒ 5,15,45 or 45,15,5.

∴ The three numbers are 5,15,45.

Let the three numbers be .

∴ According to the question

⇒ …(1)

⇒ …(2)

From 2 we get

⇒ a³ = 1728

∴ a = 12.

From 1 we get

⇒

⇒ a + ar + ar² = 65r …(3)

Substituting a = 12 in 3 we get

⇒ 12 + 12r + 12r² = 38r

⇒ 12r² – 26r + 12 = 0…(4)

Dividing (4) by 2 we get

⇒ 6r² – 13r + 6 = 0

⇒ 6r² – 12r – r + 6 = 0

⇒ 6r(r – 2) – 1(r – 2) = 0

∴ r = 2 or r = 1/2

∴ Now the equation will be

⇒

⇒ 6,12,24 or 24,12,6.

∴ The three numbers are 6,12,24.

Let the three numbers be .

∴ According to the question

⇒ …(1)

⇒ …(2)

From 2 we get

⇒ a³ = – 1

∴ a = – 1.

From 1 we get

⇒

⇒ 12a + 12ar + 12ar² = 13r …(3)

Substituting a = – 1 in 3 we get

⇒ 12( – 1) + 12( – 1)r + 12( – 1)r² = 13r

⇒12r² + 25r + 12 = 0

⇒ 12r² + 16r + 9r + 12 = 0…(4)

⇒ 4r(3r + 4) + 3(3r + 4) = 0

∴ r = or r =

∴ Now the equation will be

⇒

⇒

∴ The three numbers are .

Let the three numbers be .

∴ According to the question

⇒ …(1)

From 1 we get

⇒ a³ = 125

∴ a = 5.

⇒

⇒ a + ar + a² =

Substituting a = 5 in above equation we get

⇒ 5 + 5r + 25 =

⇒ 5r =

⇒

⇒

⇒

∴ are the three numbers.

Let the three numbers be .

∴ According to the question

⇒ …(1)

⇒ …(2)

From 2 we get

⇒ a³ = 1

∴ a = 1.

From 1 we get

⇒

⇒ 10a + 10ar + 10ar² = 39r …(3)

Substituting a = 1 in 3 we get

⇒ 10(1) + 10(1)r + 10(1)r² = 39r

⇒10r² – 29r + 10 = 0

⇒ 10r² – 25r – 4r + 10 = 0…(4)

⇒ 5r(2r – 5) – 2(2r – 5) = 0

∴ r = or r =

∴ Now the equation will be

⇒

⇒

∴ The three numbers are.

Let the three numbers be .

∴ According to the question

⇒

⇒ a + ar + ar² = 14r…(1)

First two terms are increased by 1, and third decreased by 1

∴

The above sequence is in AP.

We know in AP.

2b = a + c

∴

⇒

⇒ 2ar + 2r = ar² + a

⇒ ar² – 2ar + a = 2r …(2)

Dividing 1 by 2 we get

⇒

⇒

⇒ 1 + r + r² = 7r² – 14r + 7

⇒ 6r² – 15r – 6 = 0

⇒ 6r² – 12r – 3r – 6 = 0

⇒ 6r(r – 2) – 3(r – 2) = 0

⇒ (6r – 3) (r – 2) = 0

⇒ r = 2 or r = 1/2.

Substituting r = 2 in 2 we get

⇒ a(2)² – 2a(2) + a = 2(2)

⇒ 4a – 4a + a = 4

⇒ a = 4

Substituting r = 1/2 in 2 we get

⇒ a(1/2)² – 2a(1/2) + a = 2(1/2)

⇒ a = 4

∴ substituting a and r we get the numbers as 2,4,8.

Let the three numbers be .

∴ According to the question

⇒ …(1)

⇒ a³ = 216

⇒ a = 6

2,8,6 is added to them

∴

The above sequence is in AP.

We know in AP.

2b = a + c

⇒

Substituting a = 6 in above equation we get,

⇒

⇒

⇒ 28r = 6 + 6r² + 8r

⇒ 6r² – 20r + 6 = 0

⇒ 6r² – 18r – 2r – 6 = 0

⇒ 6r(r – 3) – 2(r – 3) = 0

⇒ (6r – 2) (r – 3) = 0

⇒ r = 3 or r = 1/3 .

∴ Now the equation will be

⇒

⇒ 2,6,18 or 18,6,2

∴ The three numbers are 2,6,18.

Let the three numbers be .

∴ According to the question

⇒ …(1)

From 1 we get

⇒ a³ = 729

∴ a = 9.

⇒

⇒ a + ar + a² = 819

Substituting a = 9 in above equation we get

⇒ 9 + 9r + 81 = 819

⇒ 9r = 729

⇒ r = 81

∴ are the three numbers.

Let the three numbers be a, ar, and ar²

∴ According to the question

⇒ a + ar + ar² = 21

a(1 + r + r²) = 21

Squaring both sides we get,

a²(1 + r + r²)² = (21)²….(1)

And from the second condition,

a² + a²r² + a²r⁴ = 189

a²(1 + r² + r⁴) = 189……(2)

Dividing both the equations we get,

Cross multiplying we get,

3 + 3r + 3r² = 7r² – 7r + 7

4r² – 10r + 4 = 0

2r² – 5r + 2 = 0

Factorizing the quadratic equation such that, on multiplication, we get 4 and on the addition, we get 5. So,

2r² – (4r + r) + 2 = 0

2r(r – 2) –1(r – 2) = 0

(2r – 1)(r – 2) = 0

r = 1/2 , r = 2

Putting the value of r in equation 2 we get,

At r = 2,

a²(1 + r² + r⁴) = 189

a²(1 + 4 + 16) = 189

a²

a² = 9

a = ±3

At r = 1/2

a² = 9 × 16

a = 3 × 4 = 12

The numbers are:

1, 9, 81 or 81, 9, 1

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 2, r = 3, n = 7

∴ Substituting the above values in (1) we get

⇒

⇒ 3⁷ – 1

⇒ 2186

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 1, r = 3, n = 8

∴ Substituting the above values in (1) we get

⇒

⇒ (3⁸ – 1)/2

⇒ 3280

Common Ratio = r =

∴ Sum of GP till infinity = …(1)

⇒ a = 1,

∴ Substituting the above values in (1), we get,

⇒

⇒

⇒

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = (a² - b²), r = (a + b), n = n

∴ Substituting the above values in (1) we get

⇒

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 4, , n = 10

∴ Substituting the above values in (1) we get,

⇒

⇒

⇒

Given expression can also be written as

⇒

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ , r = 10, n = 8

∴ Substituting the above values in (1) we get

⇒

⇒

⇒

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = √2 , , n = 8

∴ Substituting the above values in (1) we get

⇒

⇒

⇒

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ , , n = 5

∴ Substituting the above values in (1) we get,

⇒

⇒

⇒

Let S_n = (x + y) + (x² + xy + y²) + (x³ + x² y + xy² + y³) + …. to n terms

Multiplying and dividing by (x – y) we get,

(x – y) S_n = (x² – y²) + x³ + x²y + xy² – x²y – xy² – y³..upto n terms

(x – y) S_{n =} (x² + x³ + x⁴+…n terms) – (y² + y³ + y⁴ +…n terms)

We know that,

Sum of GP for n terms =

We have two G.Ps in above sum, so,

(x – y)S_n

Hence,

Common Ratio =

∴ Sum of GP for n terms = …(1)

⇒ , , n = 2n

∴ Substituting the above values in (1) we get

⇒

⇒

⇒

Common Ratio =

∴ Sum of GP for n terms = …(1)

⇒ , r = 1 + i, n = n

∴ Substituting the above values in (1) we get

⇒

⇒

⇒

Common Ratio =

∴ Sum of GP for n terms = …(1)

⇒ a = 1, r = - a, n = n

∴ Substituting the above values in (1) we get

⇒

⇒

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = x³, r = x², n = n

∴ Substituting the above values in (1) we get

⇒

⇒

Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = √7 , r = √3 , n = n

∴ Substituting the above values in (1) we get

⇒

⇒

The given expression can also be written as

⇒ …(1)

⇒ [∵ ]

Now this term is in GP.

3, 9, 27…to 11 terms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(2)

⇒ a = 3, r = 3, n = 11

∴ Substituting the above values in (2) we get

⇒

⇒ 531438/2

⇒ 265719.

Now, Adding both these we will get the required solution.

∴ 22 + 265719

⇒ 265741

The given expression can also be written as

⇒ …(1)

⇒

Now this term is in GP.

2, 4, 8…to n terms

∴ Common Ratio =

∴ Sum of GP for n terms = …(2)

⇒ a = 2, r = 2, n = n

∴ Substituting the above values in (2) we get,

⇒

⇒ 2^{n + 1} – 2.

⇒

Now this term is in GP.

1, 3, 9…to n terms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(2)

⇒ a = 1, r = 3, n = n

∴ Substituting the above values in (2) we get,

⇒

⇒

Now, Adding both these we will get the required solution.

⇒ 2^{n + 1} – 2 +

⇒

⇒

Now this term is in GP.

16, 64, 256…to 10 terms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 16, r = 4, n = 10

∴ Substituting the above values in (1) we get

⇒

⇒ 5592400

Taking 5 in common we get

5(1 + 11 + 111 + ....n)

Now Multiply and Divide by 9 we get

⇒ )

⇒ )

⇒

⇒

Now First term is in GP.

10, 100, 1000…to n terms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 10, r = 10, n = n

∴ Substituting the above values in (1) we get

⇒

⇒

For the second term the summation is n.

∴

⇒ .

Taking 7 in common we get

7(1 + 11 + 111 + ....n)

Now Multiply and Divide by 9 we get

⇒ )

⇒ )

⇒

⇒

Now First term is in GP.

10, 100, 1000…to n terms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 10, r = 10, n = n

∴ Substituting the above values in (1) we get

⇒

⇒

For the second term the summation is n.

∴

⇒ .

Taking 9 in common we get

9(1 + 11 + 111 + ....n)

Now Multiply and Divide by 9 we get,

⇒ )

⇒ )

⇒

⇒

Now first term is in GP.

10, 100, 1000…to n terms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 10, r = 10, n = n

∴ Substituting the above values in (1) we get

⇒

⇒

For the second term the summation is n.

∴

⇒ .

Let

S = 0.5 + 0.55 + 0.555 + .....n terms

Taking 5 as common we get,

S = 5(0.1 + 0.11 + 0.111 + ...nterms)

Multiply and divide by 9

⇒ )

⇒ )

⇒

⇒

Now 1 + 1 + 1 + ..n = n

For 0.1 + 0.01 + 0.001 + ..nterms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 0.1, r = , n = n

∴ Substituting the above values in (1) we get

⇒

⇒

For second term the summation is n.

∴

⇒

Let

S = 0.6 + 0.66 + 0.666 + .....n terms

Taking 6 as common we get

S = 6(0.1 + 0.11 + 0.111 + ...nterms)

Multiply and divide by 9

⇒ )

⇒ )

⇒

⇒

Now 1 + 1 + 1 + ..n = n

For 0.1 + 0.01 + 0.001 + ..nterms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(1)

⇒ a = 0.1, r = , n = n

∴ Substituting the above values in (1) we get

⇒

⇒

For second term the summation is n.

∴

⇒

Given:

Sum of GP =

∴ Common Ratio = r =

a = 3

To find: Number of terms = n.

Sum of GP for n terms =

⇒

⇒

⇒

⇒

⇒

⇒

∴ n = 10.

Given:

Sum of GP = 728

∴ Common Ratio = r =

a = 2

To find: Number of terms = n.

Sum of GP for n terms =

⇒

⇒ 728 = 3ⁿ - 1

⇒ 729 = 3ⁿ

⇒ 3⁶ = 3ⁿ

∴ n = 6.

Given:

Sum of GP = 39 + 13√3

∴ Common Ratio = r =

a = √3

To find: Number of terms = n.

Sum of GP for n terms =

⇒

⇒

⇒

⇒ 26√3 = √3(3ⁿ - 1)

⇒ 27 = 3ⁿ

⇒ n = 3.

Given:

Sum of GP = 381

∴ Common Ratio = r =

a = 3

To find: Number of terms = n.

Sum of GP for n terms =

⇒

⇒ 127 = 2ⁿ - 1

⇒ 128 = 2ⁿ

⇒ 2⁷ = 2ⁿ

∴ n = 7.

Given: Common Ratio = 3

Sum of GP = 728

Sum of GP for n terms =

Last term say it be n

∴ T_n = ar^{n - 1}

⇒ 486 = a3^{n - 1}

⇒

⇒ 1458 = a.3ⁿ …(1)

⇒

⇒

⇒ 1456 = a.3ⁿ - a…(2)

Subtracting 1 from 2 we get

⇒ 1458 - 1456 = a.3ⁿ - a.3ⁿ + a

⇒ a = 2.

∴ The first term is 2.

Sum of GP for n terms =

Sum of GP of 3 terms = 125

⇒

⇒ …(1)

Sum of GP of 6 terms = 152

⇒

⇒ …(2)

Dividing 1 by 2 we get

⇒

⇒

⇒

⇒ 125r³ + 125 = 152

⇒

⇒

Nth term of GP is T_n = ar^{n- 1}

⇒ …(1)

Nth term of GP is T_n = ar^{n- 1}

⇒ …(2)

Divide (1) by (2)

⇒

⇒

⇒

Substituting in 1 we get

a = 1

Sum of GP for n terms =

a = 1, r = , n = n.

⇒

⇒

⇒

We can write the above expression as:

The given expression can also be written as

⇒ …(1)

Now for the first term is in GP.

⇒ …upto 10 terms

∴ Common Ratio = r =

∴ Sum of GP for n terms = …(2)

⇒ a = 1, , n = 10

∴ Substituting the above values in (2) we get,

⇒

⇒

⇒

Now for the second term is in GP.

⇒ …upto 10 terms

∴ Common Ratio =

∴ Sum of GP for n terms = …(2)

⇒ , , n = 10

∴ Substituting the above values in (2) we get

⇒

⇒

Total sum

⇒

Nth term of GP is T_n = ar^{n - 1}

T₅ = a.r⁴

81 = a.r⁴…(1)

Nth term of GP is T_n = ar^{n - 1}

T₂ = a.r¹

24 = a.r¹…(2)

Divide 1 by 2

⇒

⇒

⇒

Substituting r in 2 we get,

a = 16

∴ The series is 16, 24, 54, …

∴ Sum of GP for n terms = …(1)

⇒ a = 16, r = , n = 8

∴ Substituting the above values in (1) we get,

⇒

⇒

⇒ 788.125

Sum of GP for n terms S₁ =

= …(1)

Sum of GP for 2n terms S₂ =

= …(2)

Sum of GP for 3n terms S₃ = …(3)

=

Let

∴ 1, 2 and 3 becomes

S_{1 =} K(rⁿ - 1)

S_{2 =} K(r²ⁿ - 1)

S_{3 =} K(r³ⁿ - 1)

∴ S₁² + S²_{2 =} k²(rⁿ - 1)² + k²(r²ⁿ - 1)²

⁼ k²(r²ⁿ + 1 - 2.rⁿ + r⁴ⁿ + 1 - 2.r²ⁿ)

= k²(r⁴ⁿ - r²ⁿ - 2rⁿ + 2)

L.H.S = k²(r⁴ⁿ - r²ⁿ - 2rⁿ + 2)

S₁(S₂ + S₃) = K(rⁿ - 1)[ (K(r²ⁿ - 1) + K(r³ⁿ - 1))]

= K²(rⁿ - 1)[r²ⁿ + r³ⁿ - 2]

= k²(r⁴ⁿ - r²ⁿ - 2rⁿ + 2)

Hence, Proved.

_{First n terms of GP be a, ar, ar}²…, ar^{n- 1}

_{From n + 1 term,}

_{GP = ar}ⁿ, ar^{n + 1}, …, ar^{2n - 1}

Sum of GP for n terms S₁ =

Sum of GP for next terms S₂ =

∴

⇒

⇒

Hence, Proved.

Given that a and b are roots of x² – 3x + p = 0

⇒ a + b = 3 and ab = p ...(i)

It is given that c and d are roots of x² – 12x + q = 0

⇒ c + d = 12 and cd = q...(ii)

Also given that a, b, c, d are in G.P.

Let a, b, c, d be the first four terms of a G.P.

⇒ a = a, b = ar c = ar² d = ar³

Now,

∴a + b = 3

⇒ a + ar = 3

⇒ a(1 + r) = 3…(iii)

c + d = 12

⇒ ar² + ar³ = 12

⇒ ar²(1 + r) = 12.....(iv)

From (iii) and (iv) we get

3.r² = 12

⇒ r² = 4

⇒ r = ±2

Substituting the value of r in (iii) we get a = 1

⇒ b = ar = 2∴

c = ar² = 2² = 4

d = ar³ = 2³ = 8

⇒ ab = p = 2and cd = 4×8 = 32

⇒ q + p = 32 + 2 = 34 and q−p = 32−2 = 30

⇒ q + p:q−p = 34:30 = 17:15

Hence, proved.

Given:

Sum of GP

∴ Common Ratio = r

a = 3

To find: Number of terms = n.

Sum of GP for n terms =

⇒

⇒

⇒

⇒

⇒

⇒

∴ n = 10.

The number of ancestors are 2, 4, 8, 16....it is in GP
common ratio = r
a = 2
and n = 10
Sum of GP for n terms

⇒ a = 2, r = 2, n = 10

∴ Substituting the above values in (1) we get

⇒

⇒ 2(1024 - 1)

⇒ 2(1023)

⇒ 2046

S₁ = n [First term is 1, common ratio 1; so sum to n terms = 1 + 1 + 1 + - - = n]
ii) S₂ = (2ⁿ - 1)/(2 - 1) = (2ⁿ - 1)
iii) S₃ = (3ⁿ - 1)/2
iv) S₄ = (4ⁿ - 1)/3 ..
v) So, S₁ + S₂ + 2S₃ + 3S₄ + - - - - + (n - 1)Sⁿ =
= n + (2ⁿ - 1) + (3ⁿ - 1) + (4ⁿ - 1) + - - - - - - - + (nⁿ - 1)
= n + ( - 1 - 1 - 1 .... to n - 1 terms) + (2ⁿ + 3ⁿ + 4ⁿ + .... + nⁿ) =
= n - (n - 1) + (2ⁿ + 3ⁿ + 4ⁿ + .... + nⁿ) = 1 + (2ⁿ + 3ⁿ + 4ⁿ + .... + nⁿ)
= 1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ + .... + nⁿ [Proved]

Let there be n terms.

a, ar, ar², ar³, ……………..ar^{n- 2}, ar^{n- 1}

The sum of a G.P.

Odd terms of the sequence are:

a, ar², ar⁴……………..ar^{n - 2}

So sum of this series =

⇒

According to the given problem:

⇒

⇒

⇒ r + 1 = 5

⇒ r = 4.

Let a be the first term and r be the common ratio of the G.P.

Given: and

Explanding the summation we get,

a₂ + a₄ + a₆ +………….a₂₀₀ = α

ar + ar³ + ar⁵ +………..+ ar¹⁹⁹ = α

………………..(1)

Also,

a₁ + a₃ + a₅ +………+ a₁₉₉ = β

a + ar² + ar⁴ +…………+ ar¹⁹⁸ = β

………………….(2)

From equation (1) and (2), Dividing them we get,

Hence, Proved.

Let T indicate a term of the progression.
T_1, T_2, T_3, ..., T_n, ...T_2n
T₁ = 1
T₂ = a
T₃ = ca
T₄ = c.a²
T₅ = c².a²
T_k if k is even =

T_{2n =}

T_{2n =}

S_2n = 1 + a + ca + c.a² + c².a² + c².a³ .....aⁿ. c^{n - 1}
= 1 + [ a + c.a² + c².a³.... + aⁿ.c^{n - 1} ] + [ ca + c².a² + c³.a³..... + c^{n - 1}.a^{n - 1}]
The sum of a G.P. =

For a + c.a² + c².a³.... + aⁿ.c^{n - 1}

a = a, r = ca, n = n

⇒

For [ ca + c².a² + c³.a³..... + c^{n - 1}.a^{n - 1}]

a = ca, r = ca, n = n

⇒

∴ The required result

⇒

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = r =

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = 1 and r =

⇒ sum =

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = r =

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = 8 and r =

⇒ sum =

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = r =

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = and r =

⇒ sum =

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = r =

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = 10 and r =

⇒ sum =

We observe that above progression possess a common ratio, but alternatively , adjacent terms are not possessing a common ratio. So, it consists of 2 geometric progressions.

Let, S =

⇒ S =

Let us denote the two progressions with S₁ and S₂

∴ S = S₁ + S₂

S₁ =

Common ratio = r =

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = and r = 1/9

⇒ S₁ =

S₂ =

Common ratio = r =

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = and r = 1/25

⇒ S₂ =

Hence,

S =

Using the properties of exponents:

The above term can be written as

Let S = …(1)

We observe that above progression(in power of 9) possess a common ratio. So it is a geometric progression.

Let m =

Common ratio = r =

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = and r =

⇒ m =

From equation 1 we have,

S = 9^m = 9^1/2 = 3 = RHS

Hence Proved

Let, S =

Using the properties of exponents:

The above term can be written as:

⇒ …

Denoting the terms in power with x,

We have-

S = 2^x where x =

Clearly, we observe that x is neither possessing any common ratio or any common difference. But if you observe carefully you can see that numerator is possessing an AP and denominator of various terms are in GP

Many of similar problems are solved using the method of difference approach as solved below:

As x = …..Equation 1

Multiply both sides of the equation with 1/2,we have-

⇒ ….Equation 2

Subtract equation 2 from equation 1,we have:

TIP: Make groups get rid of difference in the numerator

⇒

⇒

⇒ x =

Clearly, we have a progression with common ratio = 1/2

∴ it is a Geometric progression

Sum of infinite GP = ,where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = and r =

⇒ x =

From equation 1 we have,

S = 2^x = 2¹ = 2 = RHS

Given,

S_p = 1 + r^p + r^2p + … to ∞

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = r^p and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

As, |r|<1 ⇒ |r^p|<1 if (p>1)

∴ we can use the formula for the sum of infinite GP.

⇒ S_p = ….equation 1

As, s_p = 1 – r^p + r^2p - … to ∞

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = -r^p and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

As, |r|<1 ⇒ |r^p|<1 if (p>1)

∴ we can use the formula for the sum of infinite GP.

⇒ s_p = ….equation 2

As we have to prove - s_p + S_p = 2 s_2p

From equation 1 and 2, we get-

∴ S_p + s_p =

⇒ S_p + s_p = {using (a+b)(a-b)=a²-b²}

⇒ S_p + s_p =

As S_p =

∴ following the same analogy, we have-

∴ S_p + s_p =

Hence,

S_p + s_p = 2S_2p

Let a denote the first term of GP and r be the common ratio.

We know that nth term of a GP is given by-

a_n = ar^n-1

As, a = 4 (given)

And a₅ – a₃ = 32/81 (given)

⇒ 4r⁴ – 4r² = 32/81

⇒ 4r²(r² – 1) = 32/81

⇒ r²(r² – 1) = 8/81

Let us denote r² with y

∴ 81y(y-1) = 8

⇒ 81y² – 81y - 8 = 0

Using the formula of the quadratic equation to solve the equation, we have-

y =

⇒

∴ y = 18/162 = 1/9 or y = 144/162 = 8/9

⇒ r² = 1/9 or 8/9

∴

As GP is decreasing and all the terms are positive so we will consider only those values of r which are positive and |r|<1

∴ r =

∵ Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ the sum of respective GPs are –

S₁ = {sum of GP for r = 1/3}

S₂ = {sum of GP for r = (2√2)/3}

Let,

x = 0.125125125 ….equation 1

As 125 is the repeating term, so in all such problems multiply both sides of the equation with a number such that complete repetitive part of number comes after the decimal.

∴ multiplying equation 1 with 1000 in both sides, we have –

1000x = 125.125125125… …equation 2

Subtracting equation 1 from equation 2,we get-

1000x – x = 125.125125125… - 0.125125125….

⇒ 999x = 125

∴ x = 125/999

Let,

x = 0.4233333333….. ….equation 1

As 3 is the repeating term, so in all such problems multiply both sides of the equation with a number such that complete repetitive part of number comes after the decimal.

∴ multiplying equation 1 with 100 in both sides, we have –

100x = 42.3333333333… …equation 2

Subtracting equation 1 from equation 2,we get-

100x – x = 42.3333333… - 0.423333333…

⇒ 99x = 41.91 {as letter terms gives zero only 42.33-0.42 gives result}

∴ x = 41.91/99

⇒ x = 4191/9900

Note: We can also solve these problems using geometric progression, but the above method is much simpler.

Let,

x = 0.33333333…..

x = 0.3 + 0.03 + 0.003 + …∞

⇒ x = 3(0.1 + 0.01 + 0.001 + …∞ )

⇒ x =

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/10 and first term (a) = 1/10

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for the sum of infinite GP.

⇒ x = 3×

∴ x = 1/3

Let,

x = 0.231231231231…..

x = 0.231 + 0.000231 + 0.000000231 + …∞

⇒ x = 231(0.001 + 0.00001 + 0.0000001 + …∞ )

⇒ x =

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/1000 and first term (a) = 1/1000

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for the sum of infinite GP.

⇒ x = 231×

∴ x = 231/999

Let,

x = 3.522222222 …..

x = 3.5+0.02 + 0.002 + 0.0002 + …∞

⇒ x = 3.5+2(0.01 + 0.001 + 0.0001 + …∞ )

⇒ x =

⇒ x = 3.5 + 2S

Where S =

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/10 and first term (a) = 1/100

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for the sum of infinite GP.

⇒ S =

∴ x = 3.5 + 2(1/90)

⇒ x = (35/10) + 1/45 = (315+2)/90 = 317/90

Let,

x = 0.688888888888…..

x = 0.6+0.08 + 0.008 + 0.0008 + …∞

⇒ x = 0.6+8(0.01 + 0.001 + 0.0001 + …∞ )

⇒ x =

⇒ x = 0.6 + 2S

Where S =

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/10 and first term (a) = 1/100

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for sum of infinite GP.

⇒ S =

∴ x = 0.6 + 8(1/90)

⇒ x = (6/10) + 4/45 = (54+8)/90 = 62/90

As the midpoints of the triangles are joined successively to get another term and this is being a repeatedly infinite number of terms.

So we will be having an infinite number of side length for an infinite number of triangles.

Let ΔABC represents the equilateral triangle with side 18 cm.

D,E and F are the midpoints of side AB,BC and AC respectively

And thus ΔDEF represents another equilateral triangle.

We can find the length of DE using midpoint theorem of triangles.

If the midpoint of the 2 sides of a triangle are joined,it is parallel to the third side and is equal to 1/2 of it.

∴ DE = 1/2 × 18 = 9 cm

Similarly triangle inside DEF will have side = 9/2, and so on for other triangles.

We need to find sum of perimeters of all the triangles.

Sum of Perimeter of all the triangles = P(say)

∴ P = 3×18 + 3×9 + 3×(9/2) + 3×(9/4) + …∞

⇒ P = 54 + 27 (1 + 1/2 + 1/4 +…∞ )

⇒ P = 54 + 27S

Where S = (1 + 1/2 + 1/4 +…∞ )

We observe that above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/2 and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for sum of infinite GP.

⇒ S =

∴ P = 54 + 27×2 = 54+54 = 108

∴ Sum of the perimeters of all the triangles is 108 cm

We need to find sum of Area of all the triangles.

Sum of Perimeter of all the triangles = A(say)

As the area of an equilateral triangle is given by - ,where l represents the length of side of triangle.

∴ A =

⇒ A =

⇒ A =

⇒ P = 81√3 (1 + 1/4 + 1/16 +…∞ )

⇒ P = 81√3 S’

Where S’ = (1 + 1/4 + 1/16 +…∞ )

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = 1/4 and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for the sum of infinite GP.

⇒ S’ =

∴ A = = 108√3

∴ Sum of the Area of all the triangles is 108√3 cm²

As we have the first term of GP. Let r be the common ratio.

∴ we can say that GP is 1 , r , r² , r³ … ∞

As per the condition, each term is the sum of all terms which follow it.

If a₁,a₂ , … represents first, second, third term etc

∴ we can say that:

a₁ = a₂ + a₃ + a₄ + …∞

⇒ 1 = r + r² + r³ +…∞

Note: You can take any of the cases like a₂ = a₃ + a₄ + .. all will give the same result.

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = r and first term (a) = r

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for the sum of infinite GP.

⇒ S =

⇒

⇒ r=1−r

∴ 2r=2 or r= 1/2

Hence the series is 1, 1/2, 1/4, 1/8, 1/16...............

Suppose the 1st term is a and the common ratio is r.

∴ we can say that GP looks like: a ,ar ,ar² ,…

According to question:

a + ar = 5 …equation 1

Also, a₁ = 3(a₂+a₃+a₄+…∞) {you can take any other combination}

⇒ a = 3(ar+ar²+ar³ + …∞)

⇒ 1 = 3(r + r² + r³ +…∞)

We observe that above progression possess a common ratio. So it is a geometric progression.

Common ratio = r and first term (a) = r

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ we can use the formula for the sum of infinite GP.

Therefore

⇒

1-r = 3r

r =

From equation 1:

a+ar = a(1+r) = 5.

So,

⇒

⇒ a = 4

∴ GP is (4 , 1 , 1/4 , 1/16 , ….)

Let a be the first term of GP.

Given common ratio = r

∴ we can write GP as : a ,ar ,ar² ,ar³ …

We need to proof that: each term bears a constant ratio to the sum of all terms that follow it.

Means:

Proving for each and every individual term will be a tedious and foolish job.

So we will prove this for the n^th term, and it will validate the statement for each and every term.

N^th term is given by ar^n-1.

To prove:

We know that sum of an infinite GP is given by:

Sum of infinite GP = ,where a is the first term and k is the common ratio.

∴ arⁿ + arⁿ⁺¹ + … ∞ = arⁿ(1 + r + r² +…∞)

∴ Sum =

Hence,

As the ratio is independent of the value of each and every term

And hence we say that it bears a constant ratio. Proved.

Let a be the first term, and r be the common ratio.

According to the question-

a + ar + ar² + …∞ = S

⇒ S = a(1+r+r²+…∞)

We observe that the above progression possess a common ratio. So it is a geometric progression.

Common ratio = r and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ S = …equation 1

Also, as per the question

S₁ = a² + a²r² + a²r⁴ + …∞

⇒ S₁ = a² (1+r²+r⁴+…∞)

We observe that above progression possess a common ratio. So it is a geometric progression.

Common ratio = r² and first term (a) = 1

Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ S₁ =

⇒ S₁ =

From equation 1,we have-

⇒ S₁ = ….equation 2

Dividing equation 1 by 2, we get-

⇒

⇒ (1-r)S² = (1+r)S₁

⇒ S² – S₁ = r(S² + S₁)

∴ r =

Put the value of r in equation 1 to get a.

a =

If a, b, c are in GP

⇒ .....(i)

We know,

log a – log b = {property of logarithm}

and according to equation (i)

⇒

⇒ log b – log a = log c – log b

⇒ 2 log b = log a + log c {property of arithmetic mean}

Hence they are in AP. …proved

Given:

a, b and c are in GP

∴ b² = ac {property of geometric mean}

Taking log on both sides with base m –

log_m b² = log_m ac

⇒ log_m b² = log_m a + log_m c {using property of log}

⇒ 2log_m b = log_m a + log_m c …equation 1

Note: If three numbers a,b and c are in AP,we can say that –

2b = a + c

As equation 1 matches the form above, So

⇒ log_m a, log_m b and log_m c are in AP. …(1)

Now, applying base changing formula we get

⇒ log_ab =

∴ Applying base change on 1, we get

⇒ are in A.P

Hence, proved

Let a = k + 9; b = k−6;

c = 4

Since, a, b and c are in GP, then

b² = ac {using idea of geometric mean}

⇒ (k−6)² = 4(k + 9)

⇒ k² – 12k + 36 = 4k + 36

⇒ k² – 16k = 0

⇒ k = 0 or k = 16

Let the original numbers be

a, a + d, and a + 2d

According to the question –

a + a + d + a + 2d = 15

⇒ 3a + 3d = 15 or a + d = 5

⇒ d = 5 – a

After the addition, the three numbers are:

a + 1, a + d + 3, and a + 2d + 9

they are now in GP, that is –

⇒

⇒ (a + d + 3)^{2 =} (a + 2d + 9)(a + 1)

⇒ a² + d² + 9 + 2ad + 6d + 6a = a² + a + 2da + 2d + 9a + 9

⇒ (5 – a)² – 4a + 4(5 – a) = 0

⇒ 25 + a² – 10a – 4a + 20 – 4a = 0

⇒ a² – 18a + 45 = 0

⇒ a² – 15a – 3a + 45 = 0

⇒ a(a – 15) – 3(a – 15) = 0

⇒ a = 3 or a = 15

∴ d = 5 – a

d = 5 – 3 or d = 5 – 15

d = 2 or – 10

∴ The numbers are 3,5,7 or 15,5, – 5

Let the original numbers be

a, a + d, and a + 2d

According to the question –

3a + 3d = 21 or a + d = 7.

⇒ d = 7 – a

After the addition, the three numbers are:

a, a + d – 1, and a + 2d + 1

they are now in GP, that is –

⇒

⇒ (a + d – 1)^{2 =} a(a + 2d + 1)

⇒ a² + d² + 1 + 2ad – 2d – 2a = a² + a + 2da

⇒ (7 – a)² – 3a + 1 – 2(7 – a) = 0

⇒ 49 + a² – 14a – 3a + 1 – 14 + 2a = 0

⇒ a² – 15a + 36 = 0

⇒ a² – 12a – 3a + 36 = 0

⇒ a(a – 12) – 3(a – 12) = 0

⇒ a = 3 or a = 12

∴ d = 7 – a

d = 7 – 3 or d = 7 – 12

d = 4 or – 5

∴ The numbers are 3,7,11 or 12,7,2

Let d be the common difference of AP

∴ b = a + d ; c = a + 2d.

Given: a + b + c = 18

⇒ 3a + 3d = 18 or a + d = 6.

⇒ d = 6 – a

After the addition, the three numbers are:

a + 4, a + d + 4, and a + 2d + 36

they are now in GP, that is –

⇒

(a + d + 4)^{2 =} (a + 2d + 36)(a + 4)

⇒ a² + d² + 16 + 8a + 2ad + 8d = a² + 4a + 2da + 36a + 144 + 8d

⇒ d² – 32a – 128

⇒ (6 – a)² – 32a – 128 = 0

⇒ 36 + a² – 12a – 32a – 128 = 0

⇒ a² – 44a – 92 = 0

⇒ a² – 46a + 2a – 92 = 0

⇒ a(a – 46) + 2(a – 46) = 0

⇒ a = – 2 or a = 46

As,

d = 6 –a

∴ d = 6 – ( – 2) or d = 6 – 46

d = 8 or – 40

∴ numbers are – 2, 6, 14 or 46, 6, – 34

Let the three numbers be

∴ According to the question

⇒ …(1)

⇒ a + ar + ar² = 56r

⇒

⇒ …(2)

Subtracting 1,7,21 we get,

⇒

The above numbers are in AP

If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c

∴

⇒ 2ar – 14r = a – r + ar² – 21r

⇒ ar² – 8r + a – 2ar = 0

⇒ a(r² – 2r + 1) = 8r

From (2) we know the value of a

⇒

⇒ 56(r² – 2r + 1) = 8(1 + r + r²)

⇒ 7(r² – 2r + 1) = (1 + r + r²)

⇒ 7r² – 14r + 7 = 1 + r + r²

⇒ 6r² – 15r + 6 = 0

⇒ 6r² – 12r – 3r + 6 = 0

⇒ 6r(r – 2) – 3(r – 2) = 0

⇒ r = 2 or r = 3/6 = 1/2

When r = 2 ⇒ a = 16 {using equation 1)

r = 1/2 ⇒ a = 16

∴ the three numbers are (a/r, a, ar) = (8,16,32)

Or numbers are – (32,16,8)

Now, as a,b,c are in GP.

Using the idea of geometric mean we can write –

∴ b² = ac …(1)

Put in the LHS of the given equation to be proved –

LHS = a(ac + c²) {putting b^{2 =} ac}

⇒ LHS = a²c + ac²

⇒ LHS = c(a² + ac)

Again put ac = b²

⇒ LHS = c(a² + b²) = RHS

∴ L.H.S = R.H.S

Hence proved

Now, as a,b,c are in GP.

∴ b² = ac …(1)

Put in the LHS of the given equation to be proved –

⇒ LHS =

⇒ LHS =

⇒ LHS =

⇒ LHS =

⇒ LHS = {putting b² = ac }

⇒ LHS = a³ + b³ + c³ = RHS ...(Hence Proved)

As

a, b, c are in G.P, let r be the common ratio.

Therefore,

b = ar … (1)

c = ar² … (2)

To prove:

As, LHS =

⇒ LHS =

⇒ LHS =

⇒ LHS =

As, RHS = = LHS

Clearly, LHS = RHS

Hence proved

Now, as a,b,c are in GP.

Using the idea of geometric mean we can write –

∴ b² = ac …(1)

Put in the LHS of the given equation to be proved –

⇒ LHS =

⇒ LHS =

⇒ LHS =

⇒ LHS = = RHS

Hence Proved.

As,

a, b, c are in G.P, let r be the common ratio.

Therefore,

b = ar … (1)

c = ar² … (2)

To prove: (ab + bc + cd)² = (a + 2b + 2c) (a – 2b + 2c) = a² + 4c²

As, LHS = (a + 2b + 2c) (a – 2b + 2c)

⇒ LHS = (a + 2ar + 2ar²)(a – 2ar + 2ar²)

⇒ LHS = a²(1 + 2r + 2r²)(1 – 2r + 2r²)

⇒ LHS = a² (1 + 4r² + 4r⁴ – 4r²)

⇒ LHS = a²(1 + 4r⁴)

And RHS = a² + 4a²r⁴ = a²(1 + 4r⁴)

Clearly, LHS = RHS

Hence proved

a, b, c, d are in G.P.

Let r be the common ratio.

Therefore,

b = ar …(1)

c = ar² …(2)

and d = ar³ …(3)

If somehow we use LHS and Make it equal to RHS, our job will be done.

we can manipulate the LHS of the given equation as –

⇒ LHS =

Put the values of a,b,c and d from equation 1,2 and 3

⇒ LHS =

⇒ LHS =

⇒ LHS =

Multiplying a in numerator and denominator –

⇒ LHS =

Again from equation 1, 2, and 3, we can see –

LHS = = RHS …hence proved

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

If somehow we use RHS and Make it equal to LHS, our job will be done.

we can manipulate the RHS of the given equation as –

Note: Here we are manipulating RHS because working with a simpler algebraic equation is easier and this time RHS is looking simpler.

RHS = (a + b)² + 2(b + c)² + (c + d)²

⇒ RHS = a² + b² + 2ab + 2(c² + b² + 2cb) + c² + d² + 2cd

⇒ RHS = a² + b² + c² + d² + 2ab + 2(c² + b² + 2cb) + 2cd

Put c² = bd and b² = ac, we get –

⇒ RHS = a² + b² + c² + d² + 2(ab + ad + ac + cb + cd)

You can visualize the above expression by making separate terms for (a + b + c)² + d² + 2d(a + b + c) = {(a + b + c) + d}²

⇒ RHS = (a + b + c + d)² = LHS

Hence Proved.

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

LHS = b² + bd + bc + cd

⇒ LHS = ac + bd + bc + cd {on substituting value of b² } …(1)

RHS = c² + cd + ac + ad

⇒ RHS = bd + cd + ac + bc {putting value of c²} …(2)

From equation 1 and 2 we can say that –

LHS = RHS Hence proved

As a, b, c are in G.P.

Therefore

b² = ac … (1)

We have to prove a², b², c² are in GP or

we need to prove: (b²)² = (ac)² {using idea of GM}

On squaring equation 1 we get,

⇒ b⁴ = a²c²

⇒ (b²)² = (ac)²

Hence a²,b²,c² are in GP.

As a, b, c are in G.P.

Therefore

b² = ac … (1)

We have to prove a³, b³, c³ are in GP or

we need to prove: (b³)² = (a³c³) {using idea of GM}

On cubing equation 1 we get,

⇒ b⁶ = a³c³

⇒ (b³)² = (a³c³)

Hence a³,b³,c³ are in GP.

a, b, c are in G.P

Therefore

b² = ac … (1)

We have to prove a² + b^2, ab + bc, b² + c² are in GP or

we need to prove: (ab + bc)² = (a² + b²).(b² + c²) {using GM}

Take LHS and proceed:

⇒ LHS = (ab + bc)^{2 =} a²b² + 2ab²c + b²c²

∵ b² = ac

⇒ LHS = a²b² + 2b²(b²) + b²c²

⇒ LHS = a²b² + 2b⁴ + b²c²

⇒ LHS = a²b² + b⁴ + a²c² + b²c² {again using b² = ac }

⇒ LHS = b²(b² + a²) + c²(a² + b²)

⇒ LHS = (a² + b²)(b² + c²) = RHS

Hence a² + b², ab + bc, b² + c² are in GP.

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

To prove: (a² + b²), (b² + c²), (c² + d²) are in G.P, we need to prove that:

(a² + b²) (c² + d²) = (b² + c²)² {deduced using GM relation}

∴ RHS = (b² + c²)² = b⁴ + c⁴ + 2b²c²

= a²c² + b²d² + a²d² + b²c² {using equation 2 and 3}

⁼ c²(a² + b²) + d²(a² + b²)

= (a² + b²) (c² + d²) = LHS

∴ (a² + b²), (b² + c²), (c² + d²) are in G.P

Hence proved.

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

To prove: (a² – b²), (b² – c²), (c² – d²) are in G.P, we need to prove that:

(a² – b²) (c² – d²) = (b² – c²)² {deduced using GM relation}

∴ RHS = (b² – c²)² = b⁴ + c⁴ – 2b²c²

= a²c² + b²d² – a²d² – b²c² {using equation 2 and 3}

⁼ c²(a² – b²) – d²(a² – b²)

= (a² – b²) (c² – d²) = LHS

∴ (a² – b²), (b² – c²), (c² – d²) are in G.P

Hence proved.

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

To prove: are in G.P, we need to prove that:

{deduced using GM relation}

Or, (b² + c²)² = (a² + b²)(c² + d²)

Take LHS and proceed to prove –

LHS = (b² + c²)² = b⁴ + c⁴ + 2b²c²

= a²c² + b²d² + a²d² + b²c² {using equation 2 and 3}

⁼ c²(a² + b²) + d²(a² + b²)

= (a² + b²) (c² + d²) = RHS

∴ are in GP

Hence Proved.

As,

a, b, c, d are in G.P, let r be the common ratio.

Therefore,

b = ar … (1)

c = ar² … (2)

d = ar³ … (3)

If we show that: (ab + bc + cd)² = (a² + b² + c²) (b² + c² + d²)

we can say that:

(a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P

As, (ab + bc + cd)² = (a²r + a²r³ + a²r⁵)²

⇒ (ab + bc + cd)² = a⁴r²(1 + r² + r⁴)² …(4)

As,

(a² + b² + c²)( b² + c² + d²) = (a² + a²r² + a²r⁴)(a²r² + a²r⁴ + a²r⁶)

⇒ (a² + b² + c²)( b² + c² + d²) = a⁴r²(1 + r² + r⁴)(1 + r² + r⁴)

⇒ (a² + b² + c²)( b² + c² + d²) = a⁴r²(1 + r² + r⁴)² …(5)

From equation 4 and 5, we have:

(ab + bc + cd)² = (a² + b² + c²)(b² + c² + d²)

Hence,

We can say that (a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P.

Given as (a – b), (b – c), (c – a) are in G.P

∴

⇒ (b – c)² = (a – b)(c – a)

As we have to prove :(a + b + c)² = 3(ab + bc + ca) so we proceed as follows:

⇒ b² + c² – 2bc = ac – a² – bc + ab

⇒ a² + b² + c² = ac + ab + bc

Add 2(ac + ab + bc) to both sides:

⇒ a² + b² + c² + 2(ac + ab + bc) = ac + ab + bc + 2(ac + ab + bc)

⇒ (a + b + c)² = 3(ab + bc + ca)

Hence Proved.

As a,b,c are in GP

Note:

1. In general, the GP series is like a,ar,ar²……….

2. In this, b = ar and c = br = ar²

So we proceed forward with the aim to equalize LHS and RHS of the equation to be proved using the above ideas.

L.H.S =

⇒ LHS =

⇒ LHS =

Now

R.H.S =

⇒ RHS =

∴ RHS = 1/r

Clearly we observed that,

LHS = RHS = (1/r) …Proved

Let first term of GP be a and common ratio be r

As n^th term of GP is given as –

T_n = ar^{n– 1}

∴ T₄ = ar^{4 – 1} = ar³

Similarly T₁₀ = ar⁹

And T₁₆ = ar¹⁵

∴ x = ar³, y = ar⁹ & z = ar¹⁵

Clearly we observed that x, y, z have a common ratio.

∴ x,y,z are in GP with common ratio r⁶.Hence proved.

a, b, c are in AP

So, 2b = a + c …(1)

b, c, d are in GP

So, b² = ad …(2)

Multiply first equation with a and subtract it from 2nd.

b² – 2ab = ad – ac – a²

a² + b² – 2ab = a(d – c)

⇒ (a – b)² = a(d – c)

As a, (a – b), (d – c) satisfy the geometric mean relationship

Hence a, (a – b),(d – c) are in G.P.

Given,

pth, qth rth and sth terms of an AP are in GP .

Firstly we should find out pth, qth, rth and sth terms

Let a is the first term and d is the common difference of an AP

so, pth term = a + (p – 1)d

qth term = a + (q – 1)d

rth term = a + (r – 1)d

sth term = a + (s – 1)d

∴ [a + (p – 1)d ], [a + (q – 1)d ], [a + (r – 1)d ], [a + (s – 1)d ] are in GP

so, Let first term of GP be α and common ratio is β

Then, [a + (p – 1)d ] = α

[a + (q – 1)d ] = αβ

[a + (r – 1)d ] = αβ²

[a + (s – 1)d ] = αβ³

now, here, it is clear that α, αβ, αβ², αβ³ are in GP

NOTE: Using property of GP,we know that if a common term is multiplied with each number in a GP,series itself remains a GP

∴ α(1 – β), αβ(1 – β), αβ²(1 – β) are in GP

Where the first term is α(1 – β), and the common ratio is β

so, α(1 – β) = [a + (p – 1)d] – [a + (q – 1)d ] = (p – q)

∴ α(1 – β) = (p – q) ...... (1)

Similarly, αβ(1 – β) = αβ – αβ² = [a + (q – 1)d ] – [a + (r – 1)d] = (q – r)

∴ αβ(1 – β) = (q – r) …… (2)

And αβ²(1 – β) = αβ² – αβ³ = [α + (r – 1)d] – [α + (s – 1)d] = (r – s)

∴ αβ²(1 – β) = (r – s) …… (3)

From the above explanation, we got α(1 – β), αβ(1 – β), αβ²(1 – β) are in GP

∴ From equations (1), (2) and (3),

(p – q), (q – r), (r – s) are in GP .

Given are in AP.

⇒ {taking arithmetic mean – to get the relationship}

⇒

⇒ ab + ac + b² + bc = b² + bc + ab + b²

⇒ b² = ac

We know if a,b,c are consecutive terms of GP then b² = ac holds.

∴ a,b,c are in GP.

Take logs of each expression, using ln(x^a) = a ln(x) etc

ln (p*q) = ln(p) + ln(q):

Given,

x^a = x^b/2z^b/2 = z^c

Taking log on each term –

⇒ …(1)

The equality of the first and third expressions tells us that

⇒ …(2)

The second expression is equal to

⇒

⇒ {using equation 1}

∴

Divide through out by ln x

∴ a = b/2 + ab/2c

⇒ 2ac = bc + ab

Dividing the equation by abc –

⇒

From this are in GP.

Given:

a,b,c are in AP

∴ 2b = a + c …… (i)

b,c,d are in GP;

⇒ c² = bd …… (ii)

1/c, 1/d, 1/e are in AP;

⇒

⇒ …(iii)

From the above substituting for b & d in (ii) above,

⇒

⇒ c(c + e) = (a + c) e

⇒ c² + ce = ae + ce

⇒ c^{2 =} ae

Thus a, c, e are in GP

If a,b,c are in AP it follows that

a + c = 2b……..(1)

and a,x,b and b,y,c are in individual GPs which follows

x² = ab …….(2)

y² = bc ……..(3)

Adding eqn 2 and 3 we get,

x² + y² = ab + bc

= b(a + c)

= b.2b ( from eqn 1)

= 2b²

So we get x² + y² = 2b² which shows that they are in AP.

a, b, c are in AP

So, 2b = a + c …(1)

b, c, d are in GP

So, b² = ad …(2)

Multiply first equation with a and subtract it from 2nd.

b² – 2ab = ad – ac – a²

⇒ a² + b² – 2ab = a(d – c)

Hence a, (a – b), (d – c) are in G.P.

Let a be the first term of GP with r being the common ratio.

∴ b = ar …(1)

c = ar² …(2)

Given,

(a + b + c) = xb

⇒ (a + ar + ar²) = x(ar)

⇒ a(1 + r + r²) = ar

⇒ (1 + r + r²) = xr

⇒ r² + (1 – x)r + 1 = 0

As r is a real number ⇒ Both solutions are real.

So discriminant of the given quadratic equation D ≥ 0

As, D ≥ 0

⇒ (1 – x)² – 4(1)(1) ≥ 0

⇒ x² – 2x – 3 ≥ 0

⇒ (x – 1)(x – 3) ≥ 0

∴ x < – 1 or x > 3 …proved

Let the A.P. be A, A + D, A + 2 D, ... and G.P be x, xR, xR², ... then

a = A + (p – 1)D, b = A + (q – 1)D, c = A + (r – 1)D

⇒ a – b = (p – q)D

Also, b – c = (q – r)D

And, c – a = (r – p)D

Also a = p^th term of GP

∴ a = xR^{p – 1}

Similarly, b = xR^{q – 1} & c = xR^{r – 1}

Hence,

(a^{b – c}).(b^{c – a}).(c^{a – b}) = [(xR^{p – 1})^{(q – r)D}].[(xR^{q – 1})^{(r – p)D}].[(xR^{r – 1})^{(p – q)D}]

⁼ x^{(q – r + r – p + p – q)D}. R^{[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]D}

⇒ (a^{b – c}).(b^{c – a}).(c^{a – b}) = x⁰. R⁰

⇒ (a^{b – c}).(b^{c – a}).(c^{a – b}) = 1 …proved

Let the six terms be a₁, a₂, a₃, a₄, a₅, a₆.

And,

A = 27, B

Now, these 6 terms are between A and B.

∴ A, a₁, a₂, a₃, a₄, a₅, a₆, B.

Now all of them are in GP

So we now have 8 terms in GP with the first term being 27 and eighth being 1/81.

We know that, T_n = ar^n–1

Here T_n = , a = 27 and

⇒

⇒

⇒

∴ a₁ = Ar =

a₂ = Ar² =

a₃ = Ar³ =

a₄ = Ar⁴ =

a₅ = Ar⁵ =

a₆ = Ar⁶ =

∴ The six GM between 27 and 1/81 are .

Let the five terms be a₁, a₂, a₃, a₄, a₅.

And,

A = 16, B

Now, these 5 terms are between A and B.

∴ A, a₁, a₂, a₃, a₄, a₅, B.

Now all of them are in GP

So we now have 7 terms in GP with the first term being 16 and seventh being 1/4.

∴ T_n = ar^n–1

Here T_n, a = 16 and

⇒

⇒

⇒

∴ a₁ = Ar =

a₂ = Ar² =

a₃ = Ar³ =

a₄ = Ar⁴ =

a₅ = Ar⁵ =

∴ The six GM between 16 and 1/4 are .

Let the five terms be a₁, a₂, a₃, a₄, a₅.

And,

Now these 5 terms are between A and B.

∴ A, a₁, a₂, a₃, a₄, a₅, B.

Now all of them are in GP

So we now have 7 terms in GP with the first term being 32/9 and seventh being 81/2.

∴ T_n = ar^n–1

Here T_n = , a = and

⇒

⇒

⇒

∴ a₁ = Ar =

a₂ = Ar² =

a₃ = Ar³ =

a₄ = Ar⁴ =

a₅ = Ar⁵ =

∴ The six GM between 16 and 1/4 are .

(i) GM = √ab

Let a = 2 and b =8

GM = √2×8

= √16

= 4.

(ii) GM = √xy

Let x = a³b and y = ab³

GM = √ a³b × ab³

= √a⁴b⁴

= a²b².

(iii) GM = √ab

Let a = –2 and b = –8

GM = √–2×–8

= √–16

[we know that √–1 = i(iota)]

= 4i.

GM = √xy

Let X = 2 and Y =

GM =

=

∴

⇒

⇒ G.M = √ab

Given A.M=25, G.M = 20.

⇒ √ab = 20 …….(1)

⇒ …….(2)

⇒ a + b = 50

⇒ a = 50 – b

Putting the value of ‘a’ in equation (1), we get,

⇒

⇒ 50b – b² = 400

⇒ b² – 50b + 400 = 0

⇒ b² – 40b – 10b + 400 = 0

⇒ b(b – 40) – 10(b – 40) = 0

⇒ b = 40 or b = 10

⇒ If b = 40 then a = 10

⇒ If b = 10 then a = 40

Let the root of the quadratic equation be a and b.

According to the given condition,

⇒

⇒ a + b = 2A …..(1)

⇒ GM = √ab = G

= ab = G²…(2)

The quadratic equation is given by,

x²– x (Sum of roots) + (Product of roots) = 0

x² – x (2A) + (G²) = 0

x² – 2Ax + G² = 0 [Using (1) and (2)]

Thus, the required quadratic equation is x² – 2Ax + G² = 0.

Let the two numbers be a and b.

GM = √ab

According to the given condition,

a + b = 6√ab …(1)

(a + b)² = 36ab

Also,

(a – b)² = (a + b)² – 4ab

= 36ab – 4ab

= 32ab

⇒ a–b = √32ab

= 4√2ab …..(2)

Adding (1) and (2), we obtain

2a = (6 + 4√2)√ab

a = (3 + 2√2)√ab

substituting the value of a in (1), we obtain,

b =(3 – 2√2)√ab

∴

Thus, the required ratio is (3+2√2) : 3–2√2.

Let the root of the quadratic equation be a and b.

According to the given condition,

⇒

⇒ a + b = 16 …..(1)

⇒ GM = √ab = 5

= ab = 25 …(2)

The quadratic equation is given by,

x²– x (Sum of roots) + (Product of roots) = 0

x² – x (a + b) + (ab) = 0

x² – 16x + 25 = 0 [Using (1) and (2)]

Thus, the required quadratic equation is x² – 16x + 25 = 0.

⇒

⇒ GM = √ab

Given AM = 10, GM = 8.

⇒

⇒ a + b = 20

⇒ a = 20–b

⇒

⇒ 20b – b² = 64

⇒ b² – 20b + 64 = 0

⇒ b² – 16b – 4b + 64 = 0

⇒ b(b – 16) – 4(b – 16) = 0

⇒ b = 4 or b = 16

⇒ If b = 4 then a = 16

⇒ If b = 16 then a = 4.

Hence, the numbers are 4 and 16.

Let us suppose a and b are two numbers.

Let us say G is a number that is the Geometric mean of a and b

Therefore a, G and b must be in Geometric Progression or GP.

This means, common ratio = G/a = b/G

Or, G² = ab

Or, G = ?(ab)... (1)

Now, let us say G₁ , G₂ , G3 , .......Gn are n geometric means between a and b.

Which means that

a , G₁ , G₂ , G₃ ...... G_n , b form a G.P.

Note that the above GP has n+2 terms and the first term is a and the last term is b, which is also the (n+2)^th term

Hence, b = ar^n+2–1

where a is the first term.

So,

b = arⁿ⁺¹

r = (b/a)^1/n+1 ....(2)

Now the product of GP becomes

Product = G₁G₂G₃......G_n

= (ar)(ar²)(ar³).....(arⁿ)

= aⁿ r^{(1+2+3+4+.............+n)}

= aⁿ r^n(1+n)/2

Putting the value of r from equation 2 , we get

= aⁿ (b/a)ⁿ⁽1+n^)/2(n+1)

= (ab)^n/2

= (?ab)ⁿ

Now, putting the value from equation 1, we get,

Product = Gⁿ

Or, G₁G₂G₃......G_n = Gⁿ

Let the two numbers be a and b.

GM = √ab

According to the given condition,

⇒

a + b = 4√ab …(1)

(a + b)² = 16ab

Also,

(a – b)² = (a + b)² – 4ab

= 16ab – 4ab

= 12ab

⇒ a – b = 2√3ab…(2)

Adding (1) and (2), we obtain

2a = (4 + 2√3 )√ab

a = (2 + √3)√ab

substituting the value of a in (1), we obtain,

b =(2 – √3)√ab

∴

Thus, the required ratio is (2+√3) : (2–√3).

Let the numbers be a and b.

Now or 2A =a+b

Also, G₁ and G₂ are GM between a and b, then a, G₁, G₂, b are in G.P.

Let r be the common ratio.

Then, b = ar^4–1 = ar³

⇒

⇒

∴ G₁ = ar =

G₂ = ar² =

∴

a + b = 2A

Given: Fifth term of GP is 2

⇒ Let the first term be a and the common ratio be r.

∴ According to the question,

T₅ = 2

We know,

a_n = ar^n-1

a₅ = a.r^5-1

2 = ar⁴

GP = a,ar,ar²,…,ar⁸

Product required = a×ar×ar²×…×ar⁸

= a⁹.r³⁶

= (ar⁴)⁹

= (2)⁹

= 512

⇒ Let the first term be a and the common ratio be r.

∴ According to the question,

a_p+q = m.

a_p-q = n.

a_n = ar^n-1

a_p+q = a.r^p+q-1

a_p-q = a.r^p-q-1

∴ a.r^p+q-1 = m.

a.r^p-q-1 = n.

Multiplying above two equations we get

a²r^{(p+q-1+(p-q-1)} = a²r^(2p-2)

a²r^(2p-2) = m.n

(ar)^2(p-1) = m.n

∴ ar^p-1 =√m.n

⇒ P^th term is given by a.r^p-1

∴ ar^p-1 =√m.n

We know when three terms say a,b,c are in GP

We can write

b² = a.c

∴ According to the given data

We can write

(a^x/2)² = log_xa . log_bx

a^x = log_xa . log_bx

⇒

⇒ a^x = log_ba

Multiplying by log_a to both sides we get

⇒ log_a (a^x) = log_a (log_ba)

⇒ x log_aa = log_a (log_ba)

⇒ x = log_a (log_ba)

Let the given GP be a,ar,ar²,…

Sum of infinite GP is given by

∴ According to the question

⇒

⇒ a = 3(1-r)

⇒ a =3-3r

⇒ a+3r = 3…(1)

⇒

⇒ the first term is a² and the common ratio is r².

∴ According to the question

⇒

⇒

⇒