Find the sum of the following series to n terms:
13 + 33 + 53 + 73 + ……..
nth term would be = 2n - 1
We know,
Therefore,
………….equation 1
…………….equation 2
From equation 1
=
[replace 2n by n]
=
Substituting in equation 2
If the line segment joining the points P(x1, y1) and Q(x2, y2) subtends an angle α at the origin O, prove that : OP. OQ cos α = x1 x2 + y1 y2.
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.
Find the sum of the following series to n terms:
23 + 43 + 63 + 83 + ………
nth term would be 2n
We know ……(1)
Therefore,
Substituting the value from 1
The vertices of a triangle ABC are A(0, 0), B (2, -1) and C (9, 0). Find cos B.
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
Four points A (6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that, find x.
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(x1,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
Find the sum of the following series to n terms:
1.2.5 + 2.3.6 + 3.4.7 + ……..
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus,
=……(1)
We know
Substituting in (1)
Find the sum of the following series to n terms:
1.2.4 + 2.3.7 + 3.4.10 + …………
The nth term be n(n + 1)(3n + 1)
1.2.4 + 2.3.7 + 3.4.10 + …………=
We know by property that:
∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
…..(1)
We know
Thus from (1)
The points A (2, 0), B(9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.
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
Find the sum of the following series to n terms:
1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ………
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus,
We know,
Substituting
Thus the answer is
Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (-36, 7), (20, 7) and (0, -8).
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(x1,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
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.
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,y1 �) and Q(x2,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
Find the sum of the following series to n terms:
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ………….
The last term be n(n + 1)
The generalized equation be
……………….(1)
Since We know by property that:
∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
We know
Thus substituting in (1)
Find the distance between P(x1, y1) and Q(x2, y2) when (i) PQ is parallel to the y-axis (ii) PQ is parallel to the x-axis.
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.
Find the sum of the following series to n terms:
3 × 12 + 5 × 22 + 7 × 32 + …………..
The nth term will be n2×(2n + 1)
The generalized equation be
We know by property that:
∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus
………(1)
We know
Substituting the values in (1)
=
=
=
=
Find a point on the x-axis, which is equidistant from the point (7, 6) and (3, 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 =
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
Find the sum of the series whose nth term is :
2n3 + 3n2 – 1
1st term = 2(1)3 + 3(1)2 – 1
2nd term = 2(2)3 + 3(2)2 - 1
And so on
Nth term = 2n3 + 3n2 – 1
General term be = 2r3 + 3r2 – 1
Summation = 1st term + 2nd 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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus
……(2)
We know
Thus substituting the above values in (2)
Summation =
Find the sum of the series whose nth term is :
n3 – 3n
Generalized term be n3 – 3n
1st term = (1)3 – 3(1)
2nd term = (2)3 – 3(2)
And so on
nth term= n3 – 3n
general term= r3 – 3r
Summation=1st term + 2nd 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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus
………(2)
We know,
Since, where if
Thus substituting the above values in (2)
Summation
Find the sum of the series whose nth term is :
n(n + 1) (n + 4)
Generalized term be r(r + 1) (r + 4)
1st term = (1)((1) + 1) ((1) + 4)
2nd term =(2)((2) + 1) ((2) + 4)
And so on
nth term = n(n + 1) (n + 4) = n3 + 5n2 + 4n
Summation = 1st term + 2nd 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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus
(2)
We know,
Thus substituting the above values in (2)
Summation
Find the sum of the series whose nth term is :
(2n – 1)2
Generalized term be (2r - 1)2=4r2 + 1 - 4r
1st term = 4(1)2 + 1 - 4(1)
2nd term =4(2)2 + 1 - 4(2)
And so on
nth term= 4n2 + 1 - 4n
Summation=1st term + 2nd 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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus
(2)
We know
Thus substituting above values in (2)
Find the 20th term and the sum of 20 terms of the series :
2 × 4 + 4 × 6 + 6 × 8 + ……………..
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 20th term be 2(20) {2(20) + 2} = 40 × 42 = 1680
Summation=1st term + 2nd 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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus,
(2)
We know
Thus
Thus substituting in above equation in (2)
=4(10)(7)(41) + 4(10)(21)
=12320
Find the locus of a point equidistant from the point (2, 4) and the y-axis.
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
Sum the following series to n terms :
3 + 5 + 9 + 15 + 23 + ………….
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
(4)
We know
Thus substituting the above values in(4)
Find the equation of the locus of a point which moves such that the ratio of its distance from (2, 0) and (1, 3) is 5 : 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
Sum the following series to n terms :
2 + 5 + 10 + 17 + 26 + ………..
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
(4)
We know
Thus substituting the above values in(4)
A point moves as so that the difference of its distances from (ae, 0) and (-ae, 0) is 2a, prove that the equation to its locus is , where b2 = a2(e2 – 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 =
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
Sum the following series to n terms :
1 + 3 + 7 + 13 + 21 + …………….
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
(4)
We know
Thus substituting the above values in(4)
Find the locus of a point such that the sum of its distances from (0, 2) and (0, -2) is 6.
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
Sum the following series to n terms :
3 + 7 + 14 + 24 + 37 + ……………..
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
Sum the following series to n terms :
1 + 3 + 6 + 10 + 15 + …………….
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
…… (4)
We know
Thus substituting the above values in(4)
Find the locus of a point which is equidistant from (1, 3) and x-axis.
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
Sum the following series to n terms :
1 + 4 + 13 + 40 + 121 + ……….
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 + …….3n - 1) ………(3)
we know
Substituting the above-given value in (3)
thus
s = 1 + 4 + 13 + 40 + 121 + …………. +
We know by property that:
∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
………….. (4)
We know
Thus substituting the above values in(4)
Find the locus of a point which moves such that its distance from the origin is three times is the distance from the x-axis.
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
A(5, 3), B(3, -2) are two fixed points, find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 units.
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(x1,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
Sum the following series to n terms :
4 + 6 + 9 + 13 + 18 + ………..
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
(4)
We know
Thus substituting the above values in(4)
Find the locus of a point such that the line segments having end points (2, 0) and (-2, 0) subtend a right angle at that point.
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
Sum the following series to n terms :
2 + 4 + 7 + 11 + 16 + ………….
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
….(4)
We know
Thus substituting the above values in(4)
Sum the following series to n terms :
The general term would be
The nth term would be
If A (-1, 1) and B (2, 3) are two fixed points, find the locus of a point P so that the area d ΔPAB = 8 sq. units.
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(x1,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
A rod of length l slides between the two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.
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
Sum the following series to n terms :
The general term would be
The nth term would be
Find the locus of the mid-point of the portion of the x cos α + y sin α = p which is intercepted between the axes.
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
If O is the origin and Q is a variable point on y2 = x, Find the locus of the mid-point of OQ.
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
Write the sum of the series : 2 + 4 + 6 + 8 + ….. + 2n
Let S = 2 + 4 + 6 + 8 + ….. + 2n
S = 2(1 + 2 + 3 + 4 + ….. + n)
Substituting the above value
S = n(n + 1)
Write the sum of the series : 12 – 22 + 32 – 42 + 52 – 62 + ……. + (2n – 1)2 – (2n)2
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
Write the sum to n terms of a series whose rth term is: r + 2r
The general term be = i + 2i
= Sum of series
We know by property that:
∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus
Substituting the above value
Thus
If , find .
………….. (1)
Given
From (1) we have
Solving the above equation
n = 10
We know
………(2)
Thus
Putting n = 10 in eq(2)
= 552
= 3025
If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then write the value of k.
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
Write the sum of 20 terms of the series :
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Since ,
Thus equation (2) becomes
Write the 50th term of the series 2 + 3 + 6 + 11 + 18 + ………
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Therefore,
Since,
Thus from (3)
Nth = 2 + (n - 1)2
Hence 50th term be
50th = 2 + (50 - 1)2
50th = 2 + (49)2
Let Sn denote the sum of the cubes of first n natural numbers, and sn denote the sum of first n natural numbers. Then write the value of
To find
Let I = …… (1)
Given,
Substituting in equation (1)
We know by property that:
∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
Thus
And We know
Substituting the values
The sum to n terms of the series
A.
B.
C.
D.
To find:
Rationalizing the above equation:
since a2 - b2 = (a + b)(a - b)
The sum of the series :
A.
B.
C.
D. None of these
We know logab=logb/loga
We know logmn=nlogm
= [1 + 2 + 3…….n]/2
= n(n + 1)/4
The value of is equal to
A.
B.
C.
D. none of these
We know by property that∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
We know
(1)
(2)
(3)
Adding (1) (2) and (3)
=
If , then
A. 2870
B. 2160
C. 2970
D. none of these
Solving we get n=20
Substituting the value
10(7)(41)
2870
If , then Sn is equal to
A. 2n – n - 1
B.
C.
D. 2n– 1
1 + 2 + 22 + …….2r - 1=1(2r - 1)/2 - 1
1 + 2 + 22 + …….2r - 1=1(2r - 1)
If to n terms is S. Then, S is equal to
A.
B.
C.
D. n2
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 + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 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
The sum of 10 terms of the series is
A.
B.
C.
D.
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)
The sum of the series 12 + 32 + 52 + …… to n terms is
A.
B.
C.
D.
We know
……….(1)
….(2)
Substituting (2) in (1)
The sum of the series to n terms is
A.
B.
C.
D.
We can write
Since