Question

if sum of 99 term of AP is 198 then find the sum of its middle term​

Answer 1

Answer:

Sum of its middle term is

25(a+2)

Step-by-step explanation:

Formula used:

The n th term of the A.P a, a+d, a+2d, .......

is

$t_n=a+(n-1)d$

The sum of n terms of an A.P a, a+d, a+2d,.... is

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

Given:

Sum of 99 terms of the A.P = 198

$S_{99}=198$

$\frac{99}{2}[2a+(99-1)d]=198$ (using the second formula)

$\frac{99}{2}[2a+98d]=198$

$\frac{99}{2}(2)[a+49d]=198$

$99[a+49d]=198$

$a+49d=\frac{198}{99}$

$a+49d=2$

Middle term of the given A.P

$=t_{50}$

$=a+49d$

$=2$

Sum of its middle term

$=S_{50}$

$=\frac{50}{2}[2a+(50-1)d]$

$=25[2a+49d]$

$=25[a+(a+49d)]$

$=25[a+2]$