Question

In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms. Show that $T_{20}=-112$. Find $S_{20}$.

Answer 1

It is given that the first term of an A.P is $a=2$. Also, the sum of first five terms is one fourth the sum of the next five terms, therefore,

$S_5=\frac{1}{4}(S_{10}-S_5)\Rightarrow 4S_5=S_{10}-S_5\Rightarrow 4S_5+S_5=S_{10}\Rightarrow 5S_5=S_{10}$

We know that sum $S_n$ of $n$ terms of an A.P with first term $a$ and common difference $d$ is:

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

Thus,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 5\times \frac{5}{2}[2a+(5-1\left)d\right]=\frac{10}{2}[2a+(10-1\left)d\right]\Rightarrow 5\left[\right(2\times 2)+4d]=\frac{2}{5}\times \frac{10}{2}\left[\right(2\times 2)+9d]\Rightarrow 5(4+4d)=2(4+9d)\Rightarrow 20+20d=8+18d\Rightarrow 20d-18d=8-20\Rightarrow 2d=-12\Rightarrow d=-\frac{12}{2}\Rightarrow d=-6$

We also know that the $n$th term of an A.P with first term $a$ and common difference $d$ is $t_n=a+(n-1)d$, therefore, with $a=2$, $n=20$ and $d=-6$, we have

$t_{20}=2+(20-1)(-6)=2+(19\times -6)=2-114=-112$

Now,

$S_{20}=\frac{20}{2}\left[\right(2\times 2)+(20-1\left)\right(-6\left)\right]=10[4+(19\times -6\left)\right]=10(4-114)=10\times -110=-1100$

Hence, $T_{20}=-112$ and $S_{20}=-1100$