Question

If the sum to infinity of the series $1+4x+7{{x}^{2}}+10{{x}^{3}}+...$ is $\dfrac{35}{16}$ , then the value of x should be equal to
(a) $\dfrac{1}{5}$
(b) $\dfrac{2}{5}$
(c) $\dfrac{3}{7}$
(d) $\dfrac{1}{7}$

Answer 1

Hint: We are given a series in which each term is a product of the terms corresponding to an arithmetic progression (1,4,7,10…) with first term 1 and common difference 3 and a geometric progression $\left( {{x}^{0}},{{x}^{1}},{{x}^{2}},{{x}^{3}}... \right)$ with first term 1 and common ratio x. Thus, the given series is in the form of an Arithmetic-geometric series. Therefore, we can use the formula for the sum of the terms of an arithmetic-geometric series and then equate it to $\dfrac{35}{16}$ and solve the equation to obtain the value of x.

Complete step-by-step solution -
The given series is $1+4x+7{{x}^{2}}+10{{x}^{3}}+...$ . We find that we can write the series in the form
$1+4x+7{{x}^{2}}+10{{x}^{3}}+...={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}+{{a}_{4}}{{b}_{4}}+...$
Where ${{a}_{1}}=1,{{a}_{2}}=1+3=4,{{a}_{3}}=4+3=7...$ and ${{b}_{1}}={{x}^{0}}=1,{{b}_{2}}=1\times x=x,{{b}_{3}}=x\times x={{x}^{2}}....$ …………….(1.1)
We see that as the next term in ${{a}_{n}}$ is obtained by adding 3 to the previous term. Therefore, ${{a}_{n}}$ is an arithmetic progression. Also, as next term of ${{b}_{n}}$ is obtained by multiplying the previous term by x, it is in the form of a geometric progression with common ratio x………………(1.2)
Also, we know an arithmetic-geometric progression is given by
${{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}+...$
Where ${{a}_{n}}$ is an arithmetic progression and ${{b}_{n}}$ is a geometric progression………..(1.3)
The sum of all the terms in an arithmetic-geometric progression where the first term in the A.P. is ${{a}_{1}}$ and the common difference is d and the first term of the G.P. is ${{b}_{1}}$ and the common difference is r is given by the formula
$\dfrac{{{a}_{1}}{{b}_{1}}}{1-r}+\dfrac{d{{b}_{1}}r}{{{\left( 1-r \right)}^{2}}}$
if r<0 and the sum is infinity if $r\ge 1$ ……………………..(1.4)
Therefore, from (1.1), (1.2) and (1.3) we see that the given series is in an arithmetic-geometric progression as the sum of its terms is given to be $\dfrac{35}{16}$ , from (1.4), we obtain
$\begin{align}
  & \dfrac{1\times 1}{1-x}+\dfrac{3\times 1\times x}{{{\left( 1-x \right)}^{2}}}=\dfrac{35}{16} \\
 & \Rightarrow \dfrac{1-x+3x}{{{\left( 1-x \right)}^{2}}}=\dfrac{35}{16} \\
 & \Rightarrow 16\left( 1+2x \right)=35{{\left( 1-x \right)}^{2}} \\
\end{align}$
We can expand the RHS using the formula ${{\left( 1-x \right)}^{2}}=1+{{x}^{2}}-2x$ to obtain
$\begin{align}
  & 16(1+2x)=35\left( 1+{{x}^{2}}-2x \right) \\
 & \Rightarrow 16+32x=35+35{{x}^{2}}-70x \\
 & \Rightarrow 35{{x}^{2}}-102x+19=0.........................(1.5) \\
\end{align}$
We know that the solution of a quadratic equation given by $a{{x}^{2}}+bx+c=0$ is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Using this in (1.5), we obtain
$\begin{align}
  & x=\dfrac{-\left( -102 \right)\pm \sqrt{{{\left( -102 \right)}^{2}}-4\times 35\times 19}}{2\times 35}=\dfrac{102\pm 88}{70} \\
 & \Rightarrow x=\dfrac{102+88}{70}=\dfrac{190}{70}\text{ or }x=\dfrac{102-88}{70}=\dfrac{14}{70}=\dfrac{1}{5} \\
\end{align}$
However, as the sum is given to have a finite value, using (1.4), the value of $\dfrac{190}{11}$ should be rejected as it is greater than 1.
Thus the value of x is $\dfrac{1}{5}$ which matches option (a). Hence is the correct answer.

Note: We should note that in this question, we obtained two values of x and then rejected one value as it was greater than one. However, we should always justify the reason for the rejection of one value of x if it is to be rejected else both values of x should be considered to be the correct answer.