Question

Sum of infinity of the series $1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+...$ is?
(a) $\dfrac{7}{16}$
(b) $\dfrac{5}{16}$
(c) $\dfrac{104}{64}$
(d) $\dfrac{35}{16}$

Answer 1

Assume the sum of the given arithmetic – geometric progression as S and assume the equation as (1). Find the common ratio of the GP by considering the terms $1$, $\dfrac{1}{5}$, $\dfrac{1}{{{5}^{2}}}$, $\dfrac{1}{{{5}^{3}}}$, … to be in GP. Now, multiply both the sides with the obtained common ratio and assume the equation as (2). Subtract equation (2) form (1) and use the formula of sum of infinite terms of a GP given as ${{s}_{\infty }}=\dfrac{a}{1-r}$, where a is the first term, r is the common ratio and $\left| r \right|<1$.

Complete step by step solution:
Here we have been provided with the sum $1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+...$ and we are asked to find its value if the terms goes on to infinity. Let us assume the given sum as S, so we have,
$\Rightarrow S=1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+...$ …… (1)
Clearly we can see that the numerators 1, 4, 7, 10, … are in AP and the denominators $1$ , $\dfrac{1}{5}$, $\dfrac{1}{{{5}^{2}}}$, $\dfrac{1}{{{5}^{3}}}$, … are in GP, so the given terms of the sum form an arithmetic – geometric progression (AGP). The common ratio (r) of the terms in GP can be found by dividing the second term by the first term, so we get,
$\begin{aligned} & \Rightarrow r=\dfrac{\left( \dfrac{1}{5} \right)}{1} \\\ & \Rightarrow r=\dfrac{1}{5} \\\ \end{aligned}$
Multiplying both the sides of equation (1) by $r=\dfrac{1}{5}$ we get,
$\Rightarrow \dfrac{S}{5}=\dfrac{1}{5}+\dfrac{4}{{{5}^{2}}}+\dfrac{7}{{{5}^{3}}}+\dfrac{10}{{{5}^{4}}}+...$ …….. (2)
Subtracting equation (2) form (1) we get,
$\Rightarrow \left( S-\dfrac{S}{5} \right)=\left( 1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+... \right)-\left( \dfrac{1}{5}+\dfrac{4}{{{5}^{2}}}+\dfrac{7}{{{5}^{3}}}+\dfrac{10}{{{5}^{4}}}+... \right)$
Leaving the term 1 as it is and grouping the terms with same exponent of 5 in the denominators, we can write the above expression as,

& \Rightarrow \dfrac{4S}{5}=1+\left( \dfrac{4}{5}-\dfrac{1}{5} \right)+\left( \dfrac{7}{{{5}^{2}}}-\dfrac{4}{{{5}^{2}}} \right)+\left( \dfrac{10}{{{5}^{3}}}-\dfrac{7}{{{5}^{3}}} \right)+.... \\\ & \Rightarrow \dfrac{4S}{5}=1+\left( \dfrac{3}{5} \right)+\left( \dfrac{3}{{{5}^{2}}} \right)+\left( \dfrac{3}{{{5}^{3}}} \right)+.... \\\ \end{aligned}$$ Taking 3 common starting from the second term to the infinite term we get, $$\Rightarrow \dfrac{4S}{5}=1+\left[ 3\times \left\\{ \left( \dfrac{1}{5} \right)+\left( \dfrac{1}{{{5}^{2}}} \right)+\left( \dfrac{1}{{{5}^{3}}} \right)+.... \right\\} \right]$$ Clearly the terms $\dfrac{1}{5}$, $\dfrac{1}{{{5}^{2}}}$, $\dfrac{1}{{{5}^{3}}}$, … are in GP with the first term (a) = $\dfrac{1}{5}$ and the common ratio (r) = $\dfrac{1}{5}$. Also we have $\left| r \right|<1$, so using the formula for the sum of infinite terms of the GP for such as case given as ${{s}_{\infty }}=\dfrac{a}{1-r}$ we get, $$\begin{aligned} & \Rightarrow \dfrac{4S}{5}=1+\left[ 3\times \left\\{ \dfrac{\left( \dfrac{1}{5} \right)}{\left( 1-\dfrac{1}{5} \right)} \right\\} \right] \\\ & \Rightarrow \dfrac{4S}{5}=1+\left[ 3\times \dfrac{1}{4} \right] \\\ & \Rightarrow \dfrac{4S}{5}=\dfrac{7}{4} \\\ \end{aligned}$$ Solving for the value of S we get, $$\begin{aligned} & \Rightarrow S=\dfrac{7}{4}\times \dfrac{5}{4} \\\ & \therefore S=\dfrac{35}{16} \\\ \end{aligned}$$ **Hence, option (d) is the correct answer.** **Note:** Note that here the terms were given till infinity and that is why we need not to worry about the last term when we grouped certain terms leaving 1 as it is in the solution. If there are $n$ then we have to be careful while grouping because when we leave the first term ungrouped in equation (1), as a result the last term of equation (2) is also left ungrouped. The reason is that if the ${{n}^{th}}$ term of equation (1) is of the form $\dfrac{p}{{{q}^{n}}}$ then in equation (2) the last term will become $\dfrac{p}{{{q}^{n+1}}}$ after multiplying by the common ratio.