Question

If the sum to infinite of the series $1 + 4x + 7{x^2} + 10{x^3} + ......$ is $\dfrac{{35}}{{16}}$ , then find x.
(Note:$\left| x \right| < 1$ )
A.$\dfrac{1}{5}$
B.$- \dfrac{1}{5}$
C.$\dfrac{1}{4}$
D.$\dfrac{1}{3}$

Answer 1

First multiply x in the given equation and subtract the obtained equation from the given equation. On solving the series will be in G.P. so use the formula of infinite G.P. series-
$\Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$ where a =first term and r is common ratio and r<$1$. Then substitute the given value in the equation and solve for x.

Complete step-by-step answer:
Given that the sum of infinite series$1 + 4x + 7{x^2} + 10{x^3} + ......$ ${S_\infty }$ =$\dfrac{{35}}{{16}}$
We have to find the value of x.
Let us assume,
$\Rightarrow {S_\infty } = 1 + 4x + 7{x^2} + 10{x^3} + ......$ -- (i)
Now on multiplying x both sides in eq. (i) we get,
$\Rightarrow x{S_\infty } = x\left( {1 + 4{x^{}} + 7{x^2} + 10{x^3} + ......} \right)$
On multiplication we get,
$\Rightarrow x{S_\infty } = x + 4{x^2} + 7{x^3} + 10{x^4} + ......$ -- (ii)
Now on subtracting eq. (ii) from eq. (i), we get
$\Rightarrow {S_\infty } - x{S_\infty } = \left( {1 + 4x + 7{x^2} + 10{x^3} + ......} \right) - \left( {x + 4{x^2} + 7{x^3} + 10{x^4} + ......} \right)$
Now taking ${S_\infty }$ common, we get,
$\Rightarrow \left( {1 - x} \right){S_\infty } = \left( {1 + 4x + 7{x^2} + 10{x^3} + ......} \right) - \left( {x + 4{x^2} + 7{x^3} + 10{x^4} + ......} \right)$ -- (iii)
On solving the RHS of the equation we get,
$\Rightarrow \left( {1 + \left( {4x - x} \right) + \left( {7{x^2} - 4{x^2}} \right) + \left( {10{x^3} - 7{x^3}} \right) + ......} \right)$ {On taking the common terms together}
On simplifying we get,
$\Rightarrow \left( {1 + 3x + 3{x^2} + 3{x^3} + ......} \right)$
In this series we see on observing that the first term ’a’ is$3x$ and the common ratio r=x. So this series is in geometric progression and it is an infinite series so using the formula for the sum of infinite G.P. series-
$\Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$ where r<$1$
Now putting the given values we get,
$\Rightarrow 1 + \dfrac{{3x}}{{1 - x}}$ for $\left| x \right| < 1$
On substituting this value in the equation (iii), we get,
$\; \Rightarrow \left( {1 - x} \right){S_\infty } = 1 + \dfrac{{3x}}{{1 - x}}$ for $\left| x \right| < 1$
On taking LCM on the right side we get,
$\; \Rightarrow \left( {1 - x} \right){S_\infty } = \dfrac{{1 - x + 3x}}{{1 - x}} = \dfrac{{1 + 2x}}{{1 - x}}$ for $\left| x \right| < 1$
We know that ${S_\infty }$ =$\dfrac{{35}}{{16}}$ then on substituting this value in the above equation we get,
$\Rightarrow \left( {1 - x} \right)\dfrac{{35}}{{16}} = \dfrac{{1 + 2x}}{{1 - x}}$
On cross multiplication we get,
$\Rightarrow 35{\left( {1 - x} \right)^2} = 16\left( {1 + 2x} \right)$
We know that ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ .
On using this formula we get,
$\Rightarrow 35\left( {1 + {x^2} - 2x} \right) = 16 + 32x$
On simplifying we obtain a quadratic equation,
$\Rightarrow 35{x^2} - 102x + 19 = 0$
On factoring we get,
$\Rightarrow \left( {7x - 19} \right)\left( {5x - 1} \right) = 0$
On equating the multiplication terms to zero we get,
$\Rightarrow \left( {7x - 19} \right) = 0{\text{ or }}\left( {5x - 1} \right) = 0$
$\Rightarrow x = \dfrac{{19}}{7}{\text{Or x = }}\dfrac{1}{5}$
Here given that$\left| x \right| < 1$hence$x = \dfrac{{19}}{7}$ is not possible.
Hence the correct answer is $x = \dfrac{1}{5}$ which is option D.

Note: Here the series is infinite hence the formula of infinite G.P. is used. Don’t confuse its formula with the formula of finite G.P. series which is given as-
$\Rightarrow S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ where a = first term, r= common ratio and n= number of terms.