Question

The sum of an infinite G.P. is 2 and the sum of G.P. made by
the cubes of this infinite series is 24 then the second term of
infinite G.P. is :-
(1) 3
(2) -3
(3)3/2
(4)-3/2​

Answer 1

EXPLANATION.

• GIVEN

sum of an infinite G.P = 2

the sum of G. P made by the cube of this infinite

series = 24

To find second term of G. P

Formula of infinite G. P

$\bigstar\green{\boxed{\bold{s_{ \infty} \: = \frac{a}{1 - r} \: \: when \: \: 0 < r < 1}}}$

Given,

$\bold{ \frac{a}{1 - r} = 2 \: \: ....(1) }$

$\bold{ \frac{ {a}^{3} }{1 - {r}^{3} } = 24} \: \: ......(2)$

multiply equation (1) by cube

and divide equation (1)^3 ÷ (2)

we get,

=> $\bold{ \frac{ {a}^{3} }{(1 - r) {}^{3} } } \times \frac{1 - {r}^{3} }{ {a}^{3} } = \frac{ {2}^{3} }{24}$

=> $\bold{\frac{1 - {r}^{3} }{(1 - r) {}^{3} } = \frac{1}{3}}$

=> $\bold{\frac{1 + r + {r}^{2} }{(1 - r) {}^{2} } = \frac{1}{3} }$

=> $\bold{3 + 3r + 3 {r}^{2} = {r}^{2} + 1 - 2r}$

=> $\bold{{2r}^{2} + 5r + 2 = 0}$

=> $\bold{2r {}^{2} + 4r + r + 2 = 0}$

=> $\bold{2r(r + 2) + 1(r + 2) = 0}$

=> $\bold{(2r + 1)(r + 2) = 0}$

=> $\bold{r \: = \frac{ - 1}{2} \: \: and \: \: r \: = - 2}$

when r = -1/2

put r = -1/2 in equation (1) we get,

a = 3

Therefore,

sequence will written as = 3 , -3/2, 3/4 .....

second term of infinite gp = -3/2

Answer 2

GIVEN :

• Sum of an infinite G.P is 2.
• The sum of G.P. made by the cubes of this infinite series is 24.

TO FIND :

• The second term of G. P.

FORMULA :

Formula of Infinite G.P.

${\boxed{\bold{s_{ \infty} \: = \frac{a}{1 - r} \: \: when \: \: 0 < r < 1}}}$

Given that,

${ \frac{a}{1 - r} = 2 \: \: ......(i) }$

${ \frac{ {a}^{3} }{1 - {r}^{3} } = 24} \: \: ..........(ii)$

Multiply equation (i) by cube,

Divide equation (i)³ ÷ (ii)

Now,

${ \frac{ {a}^{3} }{(1 - r) {}^{3} } } \times \frac{1 - {r}^{3} }{ {a}^{3} } = \frac{ {2}^{3} }{24}$

$= > {\frac{1 - {r}^{3} }{(1 - r) {}^{3} } = \frac{1}{3}}$

$= > {\frac{1 + r + {r}^{2} }{(1 - r) {}^{2} } = \frac{1}{3} }$

$= > {3 + 3r + 3 {r}^{2} = {r}^{2} + 1 - 2r}$

$= > {{2r}^{2} + 5r + 2 = 0}$

$= > 2 {r}^{2} + 4r + r + 2 = 0$

$= > {2r(r + 2) + 1(r + 2) = 0}$

$= > {(2r + 1)(r + 2) = 0}$

$= > {r \: = \frac{ - 1}{2} \: \: and \: \: r \: = - 2}$

When r = $\frac{ - 1}{2}$

Put r = $\frac{ - 1}{2} \: in \: equation \: (i)$

We get,

a = 3

Sequence will written as = $3,\frac{ - 3}{2},\frac{3}{4}$

Answer :

Second term of infinite G.P. = $\frac{ - 3}{2}$