Question

If $S_1$ is the sum of an AP of n odd number of terms and $S_2$ the sum of terms of the series in odd places, then what is $\frac{S_1}{S_2}$?

Answer 1

$S1:=1+3+5+7+----+(2n-1)$

$[nthterm;a+(n-1)d;nthterm=1+(n-1)2=(2n-1\left)\right]$

Sum to n terms of an AP: $\frac{n}{2}1stterm+Lastterm$;

$S1=\frac{n}{2}(1+2n-1)=\left(n^2\right)$

$S2=1+5+9+13+----+(2n-3)$

[Here a = 1; d = 4; no. of terms = n/2, assuming in S1, number of terms are even]

$S2=\frac{\frac{n}{2}}{2}(1+2n-3)=\frac{n}{4}\times (2n-2)=\frac{\left(n\right)(n-1)}{2}$

So, $\frac{S1}{S2}=\frac{\left(n^2\right)}{frac\left(n\right)(n-1)2}=\frac{2n}{(n-1)}$

Now let us consider, the number of terms in S1 are odd;

Then number of terms in S2 would be $=\frac{(n+1)}{2}$

Last term here $=1+(\frac{(n+1)}{2}-1)4=1+2n+2-4=2n-1$

$S2=\frac{\frac{(n+1)}{2})}{2}\times (1+2n-1)=\frac{\left(n\right)(n+1)}{2}$

So, in this case the ratio $\frac{S1}{S2}=\frac{2n}{(n+1)}$