Question

In a G.P, the product of the first four terms is 4, and the second term is reciprocal of the fourth term. The sum of the G.P up to infinite terms is
[a] 8
[b] -8
[c] $\dfrac{8}{3}$
[d] $\dfrac{-8}{3}$

Answer 1

Hint: Assume that the first term of the G.P is a and the common ratio is r. Hence using the fact that the product of the first four terms of the G.P is 4 form an equation in a and r. Again using the fact that the 2nd term is the reciprocal of the 4th term form an equation in a and r. Solve for a and r. Hence find the first term and common ratio of the G.P. Use the fact that since the infinite sum of the G.P is defined, $0\le \left| r \right|<1$. Hence reject values of r not satisfying the condition. Use the fact that the infinite sum of a G.P with the first term as a and common ratio as r is given by ${{S}_{\infty }}=\dfrac{a}{1-r}$. Hence find the sum up to infinite terms of the G.P

Complete step-by-step answer:
Let the first term of the G.P be a and the common ratio be r.
Hence, we have
Product of first four terms $=a\times ar\times a{{r}^{2}}\times a{{r}^{3}}={{a}^{4}}{{r}^{6}}$
Given that the product of the first four terms is 4, we have
${{a}^{4}}{{r}^{6}}=4\text{ }\left( i \right)$
Also since the 2nd term is reciprocal of the fourth term, we have
$\begin{aligned} & ar\times a{{r}^{3}}=1 \\\ & \Rightarrow {{a}^{2}}{{r}^{4}}=1\text{ }\left( ii \right) \\\ \end{aligned}$
Dividing equation (ii) by the square of the first equation, we get
$\begin{aligned} & \dfrac{{{a}^{4}}{{r}^{6}}}{{{a}^{4}}{{r}^{8}}}=\dfrac{4}{1} \\\ & \Rightarrow \dfrac{1}{{{r}^{2}}}=4 \\\ \end{aligned}$
Taking reciprocals on both sides, we get
$\begin{aligned} & {{r}^{2}}=\dfrac{1}{4} \\\ & \Rightarrow r=\pm \dfrac{1}{2} \\\ \end{aligned}$
Since both $\dfrac{1}{2},\dfrac{-1}{2}\in \left[ -1,1 \right]$, both values are possible.
When $r=\dfrac{1}{2},$ we have from equation (ii)
$\begin{aligned} & {{a}^{2}}{{r}^{4}}=1 \\\ & \Rightarrow {{a}^{2}}\left( \dfrac{1}{16} \right)=1 \\\ & \Rightarrow a=\pm 4 \\\ \end{aligned}$
Similarly, when $r=-\dfrac{1}{2},a=\pm 4$
Hence, we have $\left( a,r \right)=\left( 4,\dfrac{1}{2} \right),\left( -4,\dfrac{1}{2} \right),\left( 4,-\dfrac{1}{2} \right),\left( -4,-\dfrac{1}{2} \right)$
We know that ${{S}_{\infty }}=\dfrac{a}{1-r}$
Hence, we have
${{S}_{\infty }}=\dfrac{4}{1-\dfrac{1}{2}}=8$ or ${{S}_{\infty }}=\dfrac{-4}{1-\dfrac{1}{2}}=-8$ or ${{S}_{\infty }}=\dfrac{4}{1+\dfrac{1}{2}}=\dfrac{8}{3}$ or ${{S}_{\infty }}=\dfrac{-4}{1+\dfrac{1}{2}}=\dfrac{-8}{3}$
Hence options [a], [b], [c] and [d] are correct.

Note: Alternative Solution:
Let the four terms of the G.P be $\dfrac{a}{{{r}^{3}}},\dfrac{a}{r},ar,a{{r}^{3}}$
Hence, we have
$\begin{aligned} & \dfrac{a}{{{r}^{3}}}\times \dfrac{a}{r}\times ar\times a{{r}^{3}}=4 \\\ & \Rightarrow {{a}^{4}}=4 \\\ & \Rightarrow a=\pm \sqrt{2} \\\ \end{aligned}$
Also, the 2nd term is the reciprocal of the 4th term.
Hence, we have
$\begin{aligned} & \dfrac{a}{r}\times a{{r}^{3}}=1 \\\ & \Rightarrow 2{{r}^{2}}=1 \\\ & \Rightarrow r=\pm \dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
Hence the first term of the G.P is $\pm \dfrac{\sqrt{2}}{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}}=\pm 4$ , and the common ratio is ${{r}^{2}}=\dfrac{1}{2}$
Note that we obtained only one value for the common ratio. This is because we assumed that the common ratio is ${{r}^{2}}$ and hence positive. The negative ratio can be obtained by assuming that the common ratio is $-{{r}^{2}}$ and following a similar procedure as above.
Hence, we have
First-term $=\pm 4$ and the common ratio $=\pm \dfrac{1}{2}$, which is the same as obtained above.
Hence following similar procedure above, we have all of the options [a], [b], [c] and [d] are correct.