Some Special Series

nth term would be = 2n - 1

We know,

Therefore,

………….equation 1

…………….equation 2

From equation 1

=

[replace 2n by n]

=

Substituting in equation 2

Key points to solve the problem:

• The idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

Given,

Two points P and Q subtends an angle α at the origin as shown in figure:

From the figure we can see that points O,P and Q forms a triangle.

Clearly in ΔOPQ we have:

{from cosine formula in a triangle}

⇒ …..equation 1

From distance formula we have-

OP =

As, coordinates of O are (0, 0) ⇒ x₂ = 0 and y₂ = 0

Coordinates of P are (x₁, y₁) ⇒ x₁ = x₁ and y₁ = y₁

Similarly, OQ

And,

∴ OP² + OQ² - PQ² =

⇒ OP² + OQ² - PQ² =

Using (a-b)² = a² + b² – 2ab

∴ OP² + OQ² - PQ² = 2x₁ x₂ + 2y₁ y₂ ….equation 2

From equation 1 and 2 we have:

⇒ OP.OQ cosα = x₁ x₂ + y₁ y₂ …Proved.

nth term would be 2n

We know ……(1)

Therefore,

Substituting the value from 1

Key points to solve the problem:

• The idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

Given,

Coordinates of the triangle and we need to find cos B which can be easily found using cosine formula.

See the figure:

From cosine formula in ΔABC , We have:

cos B =

using distance formula we have:

AB =

BC =

And, AC =

∴ cos B = …ans

Key points to solve the problem:

• The idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

•Area of a ΔPQR – Let P(x_1,y₁) , Q(x₂,y₂) and R(x₃,y₃) be the 3 vertices of ΔPQR.

Ar(ΔPQR) =

Given, coordinates of the triangle as shown in the figure.

Also,

ar(ΔDBC) =

=

Similarly, ar(ΔABC)

∴

⇒ 24.5 = 28x – 14

⇒ 28x = 38.5

⇒ x = 38.5/28 = 1.375 …ans

The nth term be n(n + 1)(n + 4)

Thus we can write 1.2.5 + 2.3.6 + 3.4.7 + ……..

The general term would be r(r + 1)(r + 4)

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus,

=……(1)

We know

Substituting in (1)

The nth term be n(n + 1)(3n + 1)

1.2.4 + 2.3.7 + 3.4.10 + …………=

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

…..(1)

We know

Thus from (1)

Key points to solve the problem:

• The idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

• The idea of Rhombus – It is a quadrilateral with all four sides equal.

Given, coordinates of 4 points that form a quadrilateral as shown in fig:

Using distance formula, we have:

AB =

BC =

Clearly, AB ≠ BC ⇒ quad ABCD does not have all 4 sides equal.

∴ ABCD is not a Rhombus …ans

The nth term be

Where = (n - 0) + (n - 1) + (n - 2) + …… + (n - n)

Since,

………(1)

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ………….=

From (1)

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ………..

Thus, solving

Solving

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus,

We know,

Substituting

Thus the answer is

Key points to solve the problem:

• The idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

•

Incentre of a triangle - Let A(x_1,y₁) , B(x₂,y₂) and C(x₃,y₃) be the 3 vertices of ΔABC and O be the centre of the circle inscribed in ΔABC

O = where a, b and c are length of sides opposite to ∠ A , ∠ B and ∠ C respectively.

Given, coordinates of vertices of the triangle as shown in figure:

We need to find the coordinates of O:

Before that, we have to find a ,b and c. We will use the distance formula to find the same.

As, a = BC =

b = AC =

and c = AB =

∴ coordinates of O =

…ans

Key points to solve the problem:

• The idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

• Equilateral triangle- triangle with all 3 sides equal.

• Coordinates of the midpoint of a line segment – Let P(x₁,y_{1 �}) and Q(x_2,y₂) be the end points of line segment PQ. Then coordinated of the midpoint of PQ is given by –

Given, an equilateral triangle with base along y axis and midpoint at (0,0)

∴ coordinates of triangle will be A(0,y₁) B(0,y₂) and C(x,0)

As midpoint is at origin ⇒ y₁+y₂ = 0 ⇒ y₁ = -y₂ …..eqn 1

Also length of each side = 2a (given)

∴ AB = ….eqn 2

∴ from eqn 1 and 2:

y₁ = a and y₂ = -a

∴ 2 coordinates are – A(0,a) and B(0,-a)

See the figure:

Clearly from figure:

DC = x

Also in ΔADC: cos 30° =

∴

Squaring both sides:

∴

∴ Coordinates of C are (√3a,0) or (-√3a,0) ….ans

The last term be n(n + 1)

The generalized equation be

……………….(1)

Since We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

We know

Thus substituting in (1)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

Given, P(x₁, y₁) and Q(x₂, y₂) are two points.

i) When PQ is parallel to the y-axis

This implies that x – coordinate is constant ⇒ x₂ = x₁

∴ from distance formula:

PQ = = |y₂ - y₁|…ans

ii) When PQ is parallel to the x-axis

This implies that y – coordinate is constant ⇒ y₂ = y₁

∴ from distance formula:

PQ = = |x₂ - x₁|…ans

Note: we take modulus because square root gives both positive and negative values, but distance is always positive so we make it positive using modulus function.

The nth term will be n²×(2n + 1)

The generalized equation be

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

………(1)

We know

Substituting the values in (1)

=

=

=

=

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

As the point is on the x-axis so y-coordinate is 0.

Let the coordinate be (x,0)

Given distance of (x,0) from (7,6) and (3,4) is same.

∴ using distance formula we have:

(x - 7)² + (0 - 6)² = (x - 3)² + (0 - 4)²

⇒x² + 49 - 14x + 36 = x² + 9 - 6x + 16

∴point on x-axis is (7.5,0) …ans

1^st term = 2(1)³ + 3(1)² – 1

2^nd term = 2(2)³ + 3(2)² - 1

And so on

Nth term = 2n³ + 3n² – 1

General term be = 2r³ + 3r² – 1

Summation = 1^st term + 2^nd term + …….. + nth term

= 2(1)³ + 3(1)² – 1 + 2(2)³ + 3(2)² - 1 + …….2n³ + 3n² – 1 …(1)

We know,

Thus

From (1) we have

Summation =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

……(2)

We know

Thus substituting the above values in (2)

Summation =

Generalized term be n³ – 3ⁿ

1^st term = (1)³ – 3⁽¹⁾

2^nd term = (2)³ – 3⁽²⁾

And so on

nth term= n³ – 3ⁿ

general term= r³ – 3^r

Summation=1^st term + 2^nd term + …… + nth term

=(1)³ – 3⁽¹⁾ + (2)³ – 3⁽²⁾ + ……… + n³ – 3ⁿ………(1)

We know

Thus

From (1) we have

Summation =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

………(2)

We know,

Since, where if

Thus substituting the above values in (2)

Summation

Generalized term be r(r + 1) (r + 4)

1^st term = (1)((1) + 1) ((1) + 4)

2^nd term =(2)((2) + 1) ((2) + 4)

And so on

nth term = n(n + 1) (n + 4) = n³ + 5n² + 4n

Summation = 1^st term + 2^nd term + …… + nth term

=(1)((1) + 1) ((1) + 4) + (2)((2) + 1) ((2) + 4)……… + n³ + 5n² + 4n ……(1)

We know,

Thus

From (1) we have

Summation =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

(2)

We know,

Thus substituting the above values in (2)

Summation

Generalized term be (2r - 1)²=4r² + 1 - 4r

1^st term = 4(1)² + 1 - 4(1)

2^nd term =4(2)² + 1 - 4(2)

And so on

nth term= 4n² + 1 - 4n

Summation=1^st term + 2^nd term + …….. + nth term

= 4(1)² + 1 - 4(1) + 4(2)² + 1 - 4(2) ………4n² + 1 - 4n ……(1)

We know ,

Thus

From (1) we have

Summation =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

(2)

We know

Thus substituting above values in (2)

Given: 2 × 4 + 4 × 6 + 6 × 8 + ……………..

The nth term would be from given series 2n × (2n + 2)

The general term would be from given series 2r × (2r + 2)

Thus 20^th term be 2(20) {2(20) + 2} = 40 × 42 = 1680

Summation=1^st term + 2^nd term + …… + 20th term

= 2 × 4 + 4 × 6 + 6 × 8 + …… 40 × 42 (1)

We know

Thus

From (1) we have

Summation =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus,

(2)

We know

Thus

Thus substituting in above equation in (2)

=4(10)(7)(41) + 4(10)(21)

=12320

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k)

As we need to maintain the same distance of (h,k) from (2,4) and y-axis.

So we select a point (0,k) on the y-axis.

From distance formula:

Distance of (h,k) from (2,4) =

Distance of (h,k) from (0,k) =

According to question both distances are same.

∴

Squaring both sides:

⇒

⇒

Replace (h,k) with (x,y)

Thus, the locus of point equidistant from (2,4) and the y-axis is-

y² - 4x - 8y + 20=0 ….ans

Let s=3 + 5 + 9 + 15 + 23 + …………. + n

By shifting each term by one

S = 3 + 5 + 9 + 15 + 23 + …………. + nth ………(1)

S = 3 + 5 + 9 + 15 + …………. + (n - 1)th + nth ……..(2)

by (1) - (2) we get

0 = 3 + 2 + 4 + 6 + 8 + …….nth - (n - 1)th - n

Nth = 3 + 2 + 4 + 6 + 8 + …….2(n - 1)th

Nth = 3 + 2(1 + 2 + 3 + 4 + …….(n - 1)th) ……….(3)

we know

Substituting the above-given value in (3)

nth = 3 + n² - n

general term = 3 + r² - r

thus

S = 3 + 5 + 9 + 15 + 23 + …………. + nth =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

(4)

We know

Thus substituting the above values in(4)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the point whose locus is to be determined to be (h,k)

Distance of (h,k) from (2,0) =

Distance of (h,k) from (1,3) =

According to the question:

Squaring both sides:

16{(h - 2)² + k² } = 25{(h - 1)² + (k - 3)² }

⇒ 16{h² + 4 - 4h + k² } = 25{h² - 2h + 1 + k² - 6k + 9}

⇒ 9h² + 9k² + 14h - 150k + 186 = 0

Replace (h,k) with (x,y)

Thus, the locus of a point which moves such that the ratio of its distance from (2, 0) and (1, 3) is 5 : 4 is –

9x² + 9y² + 14x - 150y + 186 = 0 ….ans

Let S = 2 + 5 + 10 + 17 + 26 + …………. + n

By shifting each term by one

S = 2 + 5 + 10 + 17 + 26 + …………. + nth ……..(1)

S = 2 + 5 + 10 + 17 + …………. + (n - 1)th + nth ….(2)

by (1) - (2) we get

0 = 2 + 3 + 5 + 7 + 9 + …….nth - (n - 1)th - nth

Nth = 2 + (3 + 5 + 7 + 9 + …….2r + 1) ……….(3)

Nth = 2 + (summation of first (n - 1)th term)

we know,

Substituting the above given value in (3)

nth=n² - 1 + 2

general term=r² - 1 + 2

thus

S = 2 + 5 + 10 + 17 + 26 + …………. + nth =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

(4)

We know

Thus substituting the above values in(4)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the point whose locus is to be determined be (h,k)

Distance of (h,k) from (ae,0) =

Distance of (h,k) from (-ae,0) =

According to question:

⇒

Squaring both sides:

⇒

⇒

Again squaring both sides:

⇒

⇒

∴

⇒ where b² = a²(e² – 1)

Replace (h,k) with (x,y)

Thus, the locus of a point such that the difference of its distances from (ae, 0) and (-ae, 0) is 2a:

where b² = a²(e² – 1) ….proved

Let s=1 + 3 + 7 + 13 + 21 + …………. + n

By shifting each term by one

s=1 + 3 + 7 + 13 + 21 + …………. + nth (1)

s= 1 + 3 + 7 + 13 + …………. + (n - 1)th + nth (2)

by (1) - (2) we get

0=1 + 2 + 4 + 6 + 8 + …….nth - (n - 1)th - nth

nth=1 + (2 + 4 + 6 + 8 + …….2r ) (3)

nth=1 + (summation of first (n - 1)th term)

we know

Substituting the above given value in (3)

nth=1 + n² - n

general term =1 + r² - r

thus

s=1 + 3 + 7 + 13 + 21 + …………. + nth =

We know by property that∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

(4)

We know

Thus substituting the above values in(4)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the point whose locus is to be determined to be (h,k)

Distance of (h,k) from (0,2) =

Distance of (h,k) from (0,-2) =

According to the question:

⇒

Squaring both sides:

⇒

Again squaring both sides:

⇒

⇒

Replace (h,k) with (x,y)

Thus, the locus of a point such that sum of its distances from (0,2) and (0,-2) is 6:

….proved

Let S = 3 + 7 + 14 + 24 + 37 +……….

By Shifting each term by one, we get,

S = 3 + 7 + 14 + 24 + 37 + ……+ nth term……….(1)

S = 3 + 7 + 14 + 24 +…….. + (n – 1)th term + nth term …(2)

Substracting equation 2 from equation 1 we get,

0 = 3 + 4 + 7 + 10 + 13 +……+ (nth term – (n – 1)th term) - nth term

Nth term = 3 + 4 + 7 + 10+…nth term – (n – 1)th term

We can see that 3, 4, 7,…is an A.P with first term = 3 and common difference = 3

Sum of this A.P

Therefore,

S

(n – 1)th term = a + (n – 2)d

(n – 1)th term = 3 + (n – 2)3

(n – 1)th term = 3n – 3

Therefore,

S

S

Let s = 1 + 3 + 6 + 10 + 15 + …………. + n

By shifting each term by one

S = 1 + 3 + 6 + 10 + 15 + …………. + nth …………… (1)

S = 1 + 3 + 6 + 10 + …………. + (n - 1)th + nth ….(2)

by (1) - (2) we get

0 = 1 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th - n)

Nth = 1 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th - nth)

Nth = 1 + (2 + 3 + 4 + …….r + 1) ……………(3)

Nth = 1 + (summation upto (n - 1)th term)

we know

Substituting the above-given value in (3)

thus

S =3 + 5 + 9 + 15 + 23 + …………. + nth =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

…… (4)

We know

Thus substituting the above values in(4)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k)

As we need to maintain the same distance of (h,k) from (2,4) and x-axis.

So we select a point (h,0) on the x-axis.

From distance formula:

Distance of (h,k) from (1,3) =

Distance of (h,k) from (h,0) =

According to question both distances are same.

∴

Squaring both sides:

⇒

⇒

Replace (h,k) with (x,y)

Thus, the locus of a point equidistant from (1,3) and x-axis is-

x² - 2x - 6y + 10 = 0 ….ans

Let s=1 + 4 + 13 + 40 + 121 + …………. + n

By shifting each term by one

S = 1 + 4 + 13 + 40 + 121 + …………. + nth ….(1)

S = 1 + 4 + 13 + 40 + …………. + (n - 1)th + nth …(2)

by (1) - (2) we get

0 = 1 + (3 + 9 + 27 + 81 + …….nth - (n - 1)th - n)

Nth = 1 + (3 + 3² + 3³ + 3⁴ + …….nth - (n - 1)th - nth)

Nth = 1 + (3 + 3² + 3³ + …….3^{n - 1}) ………(3)

we know

Substituting the above-given value in (3)

thus

s = 1 + 4 + 13 + 40 + 121 + …………. +

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

………….. (4)

We know

Thus substituting the above values in(4)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k)

As we need to maintain a distance of (h,k) from origin such that it is 3 times the distance from the x-axis.

So we select a point (h,0) on the x-axis.

From distance formula:

Distance of (h,k) from (0,0) =

Distance of (h,k) from (h,0) =

According to question both distances are same.

∴

Squaring both sides:

h² + k² = 9k²

⇒ h² = 8k²

Replace (h,k) with (x,y)

Thus, the locus of a point is x² = 8y² …….ans

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

Area of a ΔPQR – Let P(x_1,y₁) , Q(x₂,y₂) and R(x₃,y₃) be the 3 vertices of ΔPQR.

Ar(ΔPQR) =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k). Name the moving point to be C

Given the area of ΔABC = 9

According to question:

9 =

⇒ 18=|-10-5k+3k-9+3h+2h|

⇒ |5h-2k-19|=18

∴ 5h-2k-19=18 or 5h-2k-19= -18

⇒ 5h-2k-37=0 or 5h-2k-1=0

Replace (h,k) with (x,y)

Thus, locus of point is 5x-2y-37=0 or 5x-2y-1=0 …….ans

Let s=4 + 6 + 9 + 13 + 18 + …………. + n

shifting each term by one,

s=4 + 6 + 9 + 13 + 18 + ……………….. + nth …..(1)

s= 4 + 6 + 9 + 13 + 18 + …………. + (n - 1)th + nth …(2)

by (1) - (2) we get

0 = 4 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th - nth)

Nth = 4 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th)

Nth = 4 + (2 + 3 + 4 + …….r + 1) ………..(3)

Nth = 4 + (summation upto (n - 1)th term)

we know

Substituting the above-given value in (3)

thus

s = 4 + 6 + 9 + 13 + 18 + …………. + nth =

s = 4 + 6 + 9 + 13 + 18 + …………. + nth =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

(4)

We know

Thus substituting the above values in(4)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

• Pythagoras theorem: In right triangle ΔABC : the sum of the square of two sides is equal to the square of its hypotenuse.

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k) and name the moving point to be C.

According to a question on drawing the figure, we get a right triangle Δ ABC.

From Pythagoras theorem we have:

BC² + AC² = AB²

From distance formula:

BC =

AC =

And AB = 4

∴

⇒

⇒

⇒ 2h² + 2k² – 8 = 0

⇒ h² + k² = 4

Replace (h,k) with (x,y)

Thus, the locus of a point is x² + y² = 4 ….ans

Let S = 2 + 4 + 7 + 11 + 16 + …………. + n

By shifting each term by one

S = 2 + 4 + 7 + 11 + 16 + ……………….. + nth ….(1)

S = 2 + 4 + 7 + 11 + 16 + …………. + (n - 1)th + nth …(2)

by (1) - (2) we get

0 = 2 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th - nth)

Nth = 2 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th)

nth = 2 + (2 + 3 + 4 + …….r + 1) …………(3)

nth = 2 + (summation upto (n - 1)th term)

we know

Substituting the above-given value in (3)

thus

s=2 + 4 + 7 + 11 + 16 + …………. + nth =

s=2 + 4 + 7 + 11 + …………. + nth =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

….(4)

We know

Thus substituting the above values in(4)

The general term would be

The nth term would be

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

Area of a ΔPQR – Let P(x_1,y₁) , Q(x₂,y₂) and R(x₃,y₃) be the 3 vertices of ΔPQR.

Ar(ΔPQR) =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k). Name the moving point to be C

Given the area of ΔABC = 8

According to question:

8 =

⇒ 16=|-3+k+2k-2+h-3h|

⇒ |3k-2h-5|=16

∴ 3k-2h-5=16 or 3k-2h-5= -16

⇒ 3k-2h-21=0 or 3k-2h+11=0

Replace (h,k) with (x,y)

Thus, locus of point is 3y-2x-21=0 or 3y-2x+11=0 …….ans

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

• Idea of section formula- Let two points A(x₁,y₁) and B(x₂,y₂) forms a line segment. If a point C(x,y) divides line segment AB in the ratio of m:n internally, then coordinates of C is given as:

C =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k). Name the moving point to be C

Assume the two perpendicular lines on which rod slides are x and y-axis respectively.

Here line segment AB represents the rod of length l also ΔADB formed is a right triangle. Coordinates of A and B are assumed to be (0,b) and (a,0) respectively.

∴ a² + b² = l²…eqn 1

As, (h,k) divides AB in ratio of 1:2

∴ from section formula we have coordinate of point C as-

C ==

As, a and b are assumed parameters so we have to remove it.

∵ h = 2a/3 ⇒ a = 3h/2

And k = b/3 ⇒ b = 3k

From eqn 1:

a² + b² = l²

∴

⇒

Replace (h,k) with (x,y)

Thus, the locus of a point on the rod is: ….ans

The general term would be

The nth term would be

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

• Idea of section formula- Let two points A(x₁,y₁) and B(x₂,y₂) forms a line segment. If a point C(x,y) divides line segment AB in the ratio of m:n internally, then coordinates of C is given as:

C = when m = n =1 , C becomes the midpoint of AB and C is given as C =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k). Name the moving point to be C

Given that (h,k) is the midpoint of line x cos α + y sin α = p intercepted between axes.

So we need to find the points at which x cos α + y sin α = p cuts the axes after which we will apply the section formula to get the locus.

Put y = 0

∴ x = p/cos α ⇒ coordinates on x-axis is (p/cos α , 0). Name the point A

Similarly, Put x = 0

∴ y = p/sin α ⇒ coordinates on y-axis is (0, p/sin α ). Name this point B

As C(h,k) is the midpoint of AB

∴ coordinate of C is given by:

C =

Thus,

…equation 1

and …equation 2

Squaring and adding equation 1 and 2:

⇒

Replace (h,k) with (x,y)

Thus, the locus of a point on the rod is: ….ans

Key points to solve the problem:

• Idea of section formula- Let two points A(x₁,y₁) and B(x₂,y₂) forms a line segment. If a point C(x,y) divides line segment AB in the ratio of m:n internally, then coordinates of C is given as:

C = when m = n =1 , C becomes the midpoint of AB and C is given as C =

How to approach: To find the locus of a point we first assume the coordinate of the point to be (h, k) and write a mathematical equation as per the conditions mentioned in the question and finally replace (h, k) with (x, y) to get the locus of the point.

Let the coordinates of a point whose locus is to be determined to be (h, k). Name the moving point to be C

As, coordinate of mid point is (h,k) {by our assumption},

Let Q(a,b) be the point such that Q lies on curve y² = x

b² = a ……equation 1

According to question C is the midpoint of OQ

∵C =⇒C =

∴

Similarly,

Putting values of a and b in equation 1,we have:

Replace (h,k) with (x,y)

Thus, the locus of a point is: ….ans

Let S = 2 + 4 + 6 + 8 + ….. + 2n

S = 2(1 + 2 + 3 + 4 + ….. + n)

Substituting the above value

S = n(n + 1)

S=1² + 3² + 5² ……… + (2n – 1)² – {2² + 4² + 6² + ……. + (2n)²}

S=1² + 3² + 5² ……… + (2n – 1)² – 2²{1² + 2² + 3² + ……. + (n)²} ……..(1)

We know

………(2)

…(3)

Substituting (3) in (1), we get,

Substituting (2) in the above equation

The general term be = i + 2ⁱ

= Sum of series

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

Substituting the above value

Thus

………….. (1)

Given

From (1) we have

Solving the above equation

n = 10

We know

………(2)

Thus

Putting n = 10 in eq(2)

= 55²

= 3025

we know

……(1)

Given 2 + 4 + 6 + 8…….2n=k(1 + 3 + 5…….2n - 1)

From (1) n(n + 1)=k(1 + 3 + 5…2n - 1) …………….(2)

……. (3)

Thus substituting the values from (1) in (3), we get,

1 + 3 + 5…(2n - 1) = n(2n + 1) - n(n + 1)

1 + 3 + 5…(2n - 1) = n² …………….(4)

Substituting (4) in (2), we get,

n(n + 1) = k(n²)

k = (n + 1)/n

_{The general term would be}

…………(1)

Since,

From equation (1),we get,

Thus the general term would be,

To find (2)

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Since ,

Thus equation (2) becomes

Let s=2 + 3 + 6 + 11 + 18 + …………. + n

By shifting each term by one

S =2 + 3 + 6 + 11 + 18 + …………. + nth ………… (1)

S = 2 + 3 + 6 + 11 + 18 + …………. + (n - 1)th + nth …(2)

by (1) - (2) we get

0 = 2 + 1 + 3 + 5 + 7 + …….nth - (n - 1)th - nth

Nth = 2 + (1 + 3 + 5 + 7 + 9 + …….2r - 1) ……….(3)

Nth = 2 + (summation of first (n - 1)th term)

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Therefore,

Since,

Thus from (3)

Nth = 2 + (n - 1)²

Hence 50^th term be

50^th = 2 + (50 - 1)²

50^th = 2 + (49)²

To find

Let I = …… (1)

Given,

Substituting in equation (1)

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

And We know

Substituting the values

To find:

Rationalizing the above equation:

since a² - b² = (a + b)(a - b)

We know log_ab=logb/loga

We know logmⁿ=nlogm

= [1 + 2 + 3…….n]/2

= n(n + 1)/4

We know by property that∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

We know

(1)

(2)

(3)

Adding (1) (2) and (3)

=

Solving we get n=20

Substituting the value

10(7)(41)

2870

1 + 2 + 2² + …….2^{r - 1}=1(2^r - 1)/2 - 1

1 + 2 + 2² + …….2^{r - 1}=1(2^r - 1)

Thus the nth term would be

nth=(n + 1)/2

the general term would be

rth=(r + 1)/2

We know by property that∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

=

=

=

=

Let S = √2 + √8 + √18 + √32 + ……..

It can be written as,

S = √2(1 + 2 + 3 + …….n)

We know that,

S

S

Let S = √2 + √6 + √18 + ……….

It can also be written as,

S = √2(1 + √3 + 3 + 3√3 + …….)

Now,

1 + √3 + 3 + 3√3 + …… is a G.P with common ratio √3

Sum of the 10 terms will be given by,

S₁₀ = 121(√6 + √2)

We know

……….(1)

….(2)

Substituting (2) in (1)

We can write

Since