Question

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

Answer 1

Step-1: finding sum of first ten terms.

Let $S=$ Sum of first tem terms
$S_1=$ Sum of first five terms
$S_2=$ Sum of next five terms

As we know, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Now, sum of first ten terms is
$S=\frac{10}{2}[2\times 2+(10-1\left)d\right]$
$\Rightarrow S=5[4+9d]$
$\therefore S=20+45d$

Step $2:$ finding sum of first five terms
$S_1=\frac{5}{2}[2\times 2+(5-1\left)d\right]$
$\Rightarrow S_1=\frac{5}{2}[2\times 2+4d]$
$\therefore S_1=10+10d$

Step $3:$ finding sum of next five terms.
$S_2=S-S_1$
$\Rightarrow S_2=(20+45d)-(10+10d)$
$\Rightarrow S_2=20+45d-10-10d$
$\therefore S_2=10+35d$

Step $4:$ using given condition for finding $d$.
As given, the sum of the first five terms is one-fourth of the next five terms.
So, $S_1=\frac{1}{4}\left(S_2\right)$
$\Rightarrow 4S_1=S_2$
$\Rightarrow 4(10+10d)=10+35d$
$\Rightarrow 40+40d=10+35d$
$\Rightarrow (40-35d)=10-40$
$\Rightarrow 5d=-30$
$\therefore d=-6$

Step $5:$ finding ${20}^{th}$ term
We know that $a_n=a+(n-1)d$
$a_{20}=2+(20-1)(-6)$
$\Rightarrow a_{20}=2+19\times (-6)$
$\Rightarrow a_{20}=2-114$
$\Rightarrow a_{20}=-112$

Final answer: Hence, the ${20}^{th}$ term of the $A.P.$ is $-112$.