Chapter 8, Problem 11CT

To determine

The type of progression of the given series and hence its sum.

Answer to Problem 11CT

The given series is a geometric progression. Sum of the series is $\frac{960800}{16807}$ .

Explanation of Solution

Given: The given series is-

  ${{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}$

Concept Used: The sum of any series can be represented using the summation sign as follows-

  $a_1+a_2+a_3+\mathrm{....}+a_n={\displaystyle \sum _{k=1}^n{a}_k}$

This represents a series whose $n^{th}$ term is $a_k$ and the sequence consists of n terms which are $a_1,a_2,a_3,\mathrm{....},a_n$ , i.e.- putting in k = 1, 2, 3 …. in the $n^{th}$ term of the sequence gives the respective terms of the series. Also, for any constant c,

  ${\displaystyle \sum _{k=1}^{n}c{a}_k}=c{\displaystyle \sum _{k=1}^n{a}_k}$ , holds true $\forall c\in R$ .

A series is said to be an Arithmetic Progression (or A.P) if the difference between any two consecutive terms, namely the $k^{th}$ term and the ${(k+1)}^{th}$ is a constant, i.e.-

  $a_{k+1}-a_k=d$ , $k=1,2,3,\mathrm{...}(n-1)$

This constant d is called the common difference. Sum of n terms of any A.P having first term a and common difference d can be determined using the formula-

  $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$

A series is said to be a Geometric Progression (or G.P) if the ratio of two consecutive terms, namely $k^{th}$ term and the ${(k+1)}^{th}$ is a constant, i.e.-

  $\frac{a_{k+1}}{a_k}=r$ , $k=1,2,3,\mathrm{...}(n-1)$

This constant r is called the common ratio. Sum of n terms of any G.P having first term a and common ratio r can be determined using the formula-

  $S_n=\frac{a(r^n-1)}{r-1}$ , if $\left|r\right|>1$

  $=\frac{a(1-r^n)}{1-r}$ , if $\left|r\right|<1$

Calculations: The given series is-

  ${{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}$

  $=49{{\displaystyle \sum _{n=1}^8\left(\frac{1}{7}\right)}}^{n-1}$

Thus, the $n^{th}$ term of the corresponding sequence is-

  $a_n={\left(\frac{1}{7}\right)}^{n-1}$

Thus,

  $a_k={\left(\frac{1}{7}\right)}^{k-1}$

  $a_{k+1}={\left(\frac{1}{7}\right)}^{k+1-1}={\left(\frac{1}{7}\right)}^k$

Now,

  $\frac{a_{k+1}}{a_k}=\frac{{\left(\frac{1}{7}\right)}^k}{{\left(\frac{1}{7}\right)}^{k-1}}$

  $={\left(\frac{1}{7}\right)}^{k-(k-1)}$

  $=\frac{1}{7}$ , which is a constant.

Since the ratio of two consecutive terms is a constant hence the given series is a G.P series with common ratio $\frac{1}{7}$ .

Thus, the sum of the series can be determined as follows-

  ${{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}$

  $=49{{\displaystyle \sum _{n=1}^8\left(\frac{1}{7}\right)}}^{n-1}$

  $=49\times \left[{\left(\frac{1}{7}\right)}^0+{\left(\frac{1}{7}\right)}^1+{\left(\frac{1}{7}\right)}^2+\mathrm{...}+{\left(\frac{1}{7}\right)}^7\right]$ (Sum of first 8 terms of the G.P)

  $=49\times \frac{1\times \left[1-{\left(\frac{1}{7}\right)}^8\right]}{1-\frac{1}{7}}$ , since $a=1$ and $r=\frac{1}{7}<1$

  $=49\times \frac{7}{6}\times \left[1-\frac{1}{7^8}\right]$

  $=\frac{7^3}{6}\times \left[\frac{7^8-1}{7^8}\right]$ (Since $49=7^2$ and $a^m\times a^n=a^{m+n}$ )

  $=\frac{7^8-1}{6\times 7^5}$ (Since $\frac{a^m}{a^n}=a^{m-n}$ )

  $=\frac{5764800}{100842}$

  $=\frac{960800}{16807}$

Hence, the sum of the given series is $\frac{960800}{16807}$ .