Question

An A.P consists of 31 terms . If its 16th term is m,then the sum of all the terms of this A.P.is​

Answer 1

Step-by-step explanation:

on applying formula of sum we get following result

Answer 2

The sum of all the terms of this A.P. is 31m.

Step-by-step explanation:

We are given that an A.P consists of 31 terms and its 16th term is m.

Firstly, the nth term of an A.P. is given by the following formula;

$a_n=a + (n-1)d$ ,   where a = first term  and  d = common difference

According to the question, 16th term of an A.P. is m, i.e;

$a_1_6=a + (16-1)d$

$m=a + 15d$  ----------------- [equation 1]

Now, the sum to n terms of an A.P. is given by the following formula;

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

$S_3_1=\frac{31}{2}[2a+(31-1)d]$        {because n = 31 terms}

$S_3_1=\frac{31}{2}[2a+30d]$

$S_3_1=\frac{31}{2}\times 2[a+15d]$

$S_3_1=31m$        {using equation 1}

Hence, the sum of all the terms of this A.P. is​ 31m.