Question

In an A.P., if $S_1=T_1+T_2+T_3+......+T_n$ ($n$ odd) and $S_2=T_2+T_4+T_6+....+T_{n-1}$, then find the value of $\frac{S_1}{S_2}$ in terms of $n$.

Answer 1

First term of an A.P is a = $T_1$ and common difference = d then A.P will be

$a,a+d,a+2d,............[a+(n-1)d].......(i)

$\Rightarrow S_1=\frac{n}{2}[2a+(n-1\left)d\right]$

From (i) we can say that even terms $(T_2,T_4......)$ are aslo A.P with first term = a+d and common difference = 2d

Number of terms = $\frac{n-1}{2}$ where n is odd

$S_2=\left(\frac{\frac{n-1}{2}}{2}\right)\left\{2\right(a+d)+[\frac{(n-1)}{2}-1\left]2d\right\}$

$S_2\left(\frac{n-1}{4}\right)\{2a+2d+(n-1)d-2d\}$

$S_2=\left(\frac{n-1}{4}\right)\{2a+(n-1\left)d\right\}.........\left(ii\right)$

$\frac{S_1}{S_2}=\frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\left(\frac{n-1}{4}\right)\{2a+(n-1\left)d\right\}}$

$\frac{S_1}{S_2}=\frac{n}{2}\times \frac{4}{n-1}$

$\frac{S_1}{S_2}=\frac{2n}{n-1}$