Question

In an A.P the first term is 2 and the sum of the first five terms is one- fourth of the sum of the next- five terms. Show that 20th term is- 112

Answer 1

I think this is enough to understand

Answer 2

Answer:

Step-by-step explanation:

First Term of A.P. = a = 2

Sum of first n terms = $S_n=\frac{n}{2} (2a+(n-1)d)$

Where n is the number of terms

a is the first term

d is the common difference

We are given that the sum of the first five terms is one- fourth of the sum of the next- five terms.

So, $S_5=\frac{1}{4}(S_{10}-S_5)$

$\frac{5}{2} (2(2)+(5-1)d)=\frac{1}{4}(\frac{10}{2} (2(2)+(10-1)d)-\frac{5}{2} (2(2)+(5-1)d))$

$20+20d=\frac{1}{4}(40+90d-(20+20d))$

$80+80d=40+90d-20-20d$

$80+80d=20+70d$

$10d=-60$

$d=-6$

Formula of nth term in A.P. = $a_n=a+(n-1)d$

Substitute n 20

$a_{20}=2+(20-1)(-6)$

$a_{20}=-112$

Hence proved .