Question

If the first term of an $A.P.$ is $2$ and the sum of first terms is equal to one fourth of the sum of the next five terms, thenfind the ${20}^{th}$ term.

Answer 1

It is given that first term$a=2$

Also, sum of first five terms$=\frac{1}{4}$\times Sum of next $5$ terms.

$a_1+a_2+a_3+a_4+a_5=\frac{1}{4}(a_6+a_7+a_8+a_9+a_{10})$

Sum of first five terms$=\frac{1}{4}$Sum of $6^{th}$ and ${10}^{th}$ terms

Sum of first five terms$=\frac{1}{4}$Sum of $10-$ Sum of first five terms

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

$\Rightarrow 4S_5=S_{10}-S_5$

$\Rightarrow 5S_5=S_{10}$ ......$\left(1\right)$

Finding sum of first five terms

We know that sum of $n$ terms of an A.P$S_n=\frac{n}{2}(2a+(n-1)d)$

Put $a=2,n=5$

$S_5=\frac{5}{2}(2\times 2+(5-1)d)$

$S_5=\frac{5}{2}(4+4d)$

$S_5=10+10d$

Finding sum of first ten terms

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

Put $a=2,n=10$

$S_{10}=\frac{10}{2}(2\times 2+(10-1)d)$

$=20+45d$

From eqn$\left(1\right)$

$5S_5=S_{10}$

$\Rightarrow 5(10+10d)=20+45d$

$\Rightarrow 5d=-30$ or $d=-6$

${20}^{th}$ term of an A.P$=a_{20}=2+(20-1)(-6)=-112$