AP GP HP Questions and Answers

As we know the formula of nth term of GP = ar^n-1
9th term = ar⁸ and 5th term =ar⁴
Given that
ar⁸ = 16ar⁴
=> r⁴ = 16
=> r = 2
Given ar⁶ = 96
put value of r above
a(2)⁶ = 96
a = 96/64 = 1.5
Hence, first term of the GP is 1.5.

Arithmetic mean of 61 means there total is 61 multiplied by 2 = 122
and, Geometric Mean of 11 meaning there product will be square of 11 = 121
Clearly, X = 121

Y = 1

Thus, value of X is 121.

As we know a = 1

d = 2

n = 50 .

Then,

s_{n} =\frac{n}{2}[2a+(n-1)\times d]

S₅₀= 25 × (2 × 1 + (50 − 1) × 2)

= 25 × (2 + 49 × 2)

= 25 × (2 + 98)

= 2500.

l = a + (n − 1)d

=>100 = 1 + (n − 1) × 1·5

Now by subtracting 1 from each side:

99 = (n − 1) × 1·5

On adding 1·5 to both the side, we will get 66 = n − 1

n = 67.

Now by using formula

s_{n} =\frac{n}{2}(a+l)

s_{67} =\frac{67}{2}(1+100)

= 3383.5.

a = first term

d = common difference

Now a = 4

S_{9} = 2\times S_{6}------(i)

Formula:

S_{n} =\frac{n}{2}[2a+(n-1)\times d]

By applying (i)



=>4.5(8+8d)= 6(8+5d)

=>36+36d=48+30d

6d=12

d = 2.

Given AP 6 , 9.5 , 13,......

First Term a = 6

Common difference = 9.5-6 = 3.5

Now, formula is

s_{n} =\frac{n}{2}[2a+(n-1)\times d]

To find the sum to 22 terms of an AP,

n = 22

Thus, S = \frac{22}{2}[2\times 6 + (21)3.5]

=> S = 11[12 + 73.5]

=> S = 11 x 85.5

S = 940.5

Solution:

S_{n} = \frac{n}{2} [2a + (n-1)d]

Consider sum of first c terms,

S_{c} = \frac{c}{2} [2a + (c-1)d]

Consider sum of first e terms,

S_{e} = \frac{e}{2} [2a + (e-1)d]

Given,

S_{c} = S_{e}

\frac{c}{2} [2a + (c-1)d]= \frac{e}{2} [2a + (e-1)d]

c[2a+(c−1)d]=e[2a+(e−1)d]

2ac +cd(c-1) = 2ae +ed(e-1)

2a(c−e) + d[c(c−1)−e(e−1)] = 0

2a(c-e) + d[c^{2}-c-e^{2} + e] = 0

2a(c−e)+d[(c+e)(c−e) - (c-e)] = 0

2a(c−e)+d[(c−e) (c+e-1)] = 0

2a+d(c+e-1) = 0

Now,

S_{c+e} = \frac{c + e}{2}[2a + (c + e- 1)d]

= \frac{c+e}{2} \times 0 = 0

=0

Therefore the sum of c+e terms is 0.

X + Y = 9

x³ + y³ = 243
Let’s assume x = 3 , y = 6

So: 3³+6³=243

x² + y² = 9 + 36 = 45.

S_{48}= 4 \times S_{36}

\frac{48}{2}(2a+47d) = 4\times \frac{36}{2}(2a+35d)

4a+58d=0

or 2a+29d=0

S_{30} = \frac{30}{2}(2a+29d)

=15(0)

=0.

X + Y = 12

x³ + y³ = 243
Let’s assume x = 4 , y = 8

So: 4³+8³=576

x² + y² = 16 + 64 = 80.

If p, q, r pre in h.p then \frac{1}{p},\frac{1}{q},\frac{1}{r} will be in A.P. series and \frac{1}{q} will be the AM of \frac{1}{p},\frac{1}{r}