Question

the ratio of the sum of the cubes of an infinitely decreasing gp to the sum of its squares is 12:13. the sum of the first and second term is equal to 4/3. find the sum, first term and common ratio.

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume an infinite GP series with

• First term be a

• Common ratio be r.

As its an decreasing GP, so r < 1.

And moreover its an Infinite GP series, so - 1 < r < 1.

Now,

Let suppose infinite GP series as

$\rm :\longmapsto\:a + ar + {ar}^{2} + - - - - -$

We know,

Sum of infinite GP series is given by

$\green{ \boxed{ \bf \: S_ \infty = \dfrac{a}{1 - r} \: where \: |r| < 1}}$

Now, According to statement,

Sum of first and second term is 4/3.

$\rm :\longmapsto\:a + ar = \dfrac{4}{3} - - - - (1)$

Consider,

Sum of the cube of infinite GP series,

$\rm :\longmapsto\: {a}^{3} + {a}^{3} {r}^{3} + {a}^{3} {r}^{6} + - - - -$

So, sum of this series is

$\rm :\longmapsto\:S_1 = \dfrac{ {a}^{3} }{1 - {r}^{3} } - - - (2)$

Consider,

Sum of the squares of infinite GP series,

$\rm :\longmapsto\: {a}^{2} + {a}^{2} {r}^{2} + {a}^{2} {r}^{4} + - - - -$

So, sum of this series is

$\rm :\longmapsto\:S_2 = \dfrac{ {a}^{2} }{1 - {r}^{2} } - - - (3)$

Now,

According to statement,

$\rm :\longmapsto\:\dfrac{S_1}{S_2} = \dfrac{12}{13}$

$\rm :\longmapsto\:\dfrac{ {a}^{3} }{1 - {r}^{3} } \div \dfrac{ {a}^{2} }{1 - {r}^{2} } = \dfrac{12}{13}$

$\rm :\longmapsto\:\dfrac{ {a}^{3} }{(1 - r)(1 + r + {r}^{2} )} \times \dfrac{(1 - r)(1 + r)}{{a}^{2} } = \dfrac{12}{13}$

$\green{\boxed{ \because \sf \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}$

$\green{\boxed{ \because \sf \: {x}^{2} - {y}^{2} = (x - y)(x + y)}}$

$\rm :\longmapsto\:\dfrac{a(1 + r)}{1 + r + {r}^{2} } = \dfrac{12}{13}$

$\rm :\longmapsto\:\dfrac{a+ ar}{1 + r + {r}^{2} } = \dfrac{12}{13}$

$\rm :\longmapsto\:\dfrac{4}{3(1 + r + {r}^{2}) } = \dfrac{12}{13} \: \: \: \{ \: using \: (1) \}$

$\rm :\longmapsto\:\dfrac{1}{3(1 + r + {r}^{2}) } = \dfrac{3}{13} \: \: \:$

$\rm :\longmapsto\:9 + 9r + {9r}^{2} = 13$

$\rm :\longmapsto\: {9r}^{2} + 9r - 4 = 0$

$\rm :\longmapsto\: {9r}^{2} + 12r - 3r - 4 = 0$

$\rm :\longmapsto\:3r(3r + 4) - 1(3r + 4) = 0$

$\rm :\longmapsto\:(3r - 1)(3r + 4) = 0$

$\rm :\implies\:r = \dfrac{1}{3} \: \: \: or \: \: \: - \: \dfrac{4}{3} \: \: \: \{rejected \: as \: |r| < 1 \}$

$\bf\implies \:r = \dfrac{1}{3}$

Now, Substitute value of r in equation (1), we get

$\rm :\longmapsto\:a(1 + r)= \dfrac{4}{3}$

$\rm :\longmapsto\:a(1 + \dfrac{1}{3} )= \dfrac{4}{3}$

$\rm :\longmapsto\:a(\dfrac{3 + 1}{3} )= \dfrac{4}{3}$

$\rm :\longmapsto\:a(\dfrac{4}{3} )= \dfrac{4}{3}$

$\bf\implies \:a = 1$

Now, we have ,

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:r = \dfrac{1}{3}$

We know,

Sum of infinite GP series is

$\rm :\longmapsto\:{{ \bf \: S_ \infty = \dfrac{a}{1 - r} \: where \: |r| < 1}}$

$\rm  \:  =  \: \:\dfrac{1}{ \: \: \: 1 - \dfrac{1}{3} \: \: \: \: \: \: }$

$\rm  \:  =  \: \:\dfrac{1}{ \: \: \: \dfrac{3 - 1}{3} \: \: \: \: \: \: }$

$\rm  \:  =  \: \:\dfrac{1}{ \: \: \: \dfrac{2}{3} \: \: \: \: \: \: }$

$\rm  \:  =  \: \:\dfrac{3}{2}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace \green{ \boxed{ \bf \: S_ \infty \: = \: \dfrac{3}{2}}}$

Answer 2

Answer:

Step-by-step explanation: