Chapter 8, Problem 12CT

To determine

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

Answer to Problem 12CT

The given series is an arithmetic progression. Sum of the series is 45.

Explanation of Solution

Given: The given series is-

  ${\displaystyle \sum _{n=1}^{15}(n-5)}$

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-

  ${\displaystyle \sum _{n=1}^{15}(n-5)}$

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

  $a_n=n-5$

Thus,

  $a_k=k-5$

  $a_{k+1}=(k+1)-5=k-4$

Now,

  $a_{k+1}-a_k=(k-4)-(k-5)$

  $=k-4-k+5$

  $=1$ , which is a constant.

Since the difference of two consecutive terms is a constant hence the given series is an A.P series with common difference $1$ .

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

  ${\displaystyle \sum _{n=1}^{15}(n-5)}$

  $=(-4)+(-3)+(-2)+\mathrm{....}+10$ (Sum of first 15 terms)

  $=\frac{15}{2}\left[2\times (-4)+(15-1)\times 1\right]$ , since $a=-4$ and $d=1$

  $=\frac{15}{2}\left[-8+14\right]$ ,

  $=\frac{15}{2}\times 6$

  $=15\times 3$

  $=45$

Hence, the sum of the given series is 45.