Question

If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its (m +n) terms is zero.

Or

The sum of n terms of two arithmetic progressions are in the ratio $(3n+8)$ : $(7n+15)$. Find the ration of their 12th terms.

Answer 1

Let a be the first term and d be the common difference of the given AP. Then, ( S_m = S_n\)

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

$\Rightarrow 2ma+d(m^2-m)=2an+d(n^2-n)$

$\Rightarrow 2ma-2na+d(m^2-m)-d(n^2-n)=0\Rightarrow 2a(m-n)+d(m^2-m)-d(n^2-n)=0$

$\Rightarrow 2a(m-n)+d(m^2-n^2)-d(m-n)=0$

$\Rightarrow (m-n)(2a+(m+n-1\left)d\right)=0$

$2a+(m+n-1)d=0[\because m-n\ne 0.....(i\left)\right]$

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

$=\frac{m+n}{2}\times 0=0\left[\text{from Eq. (i)}\right]$ $\text{Hence proved.}$

OR

Let $a_1,a_2$ be the first terms and $d_1,d_2$ be the common differences of the two given AP's Then, the sum of their n terms are given by

$S_n=\frac{n}{2}[2a_1+(n-1\left)d_1\right]\text{and}{S}_n^{\prime }=\frac{n}{2}[2a_2+(n-1\left)d_2\right]$[1]

It is given that $\frac{S_n}{S_n^{\prime }}=\frac{3n+8}{7n+15}\Rightarrow \frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}=\frac{3n+8}{7n+15}$

$\Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{3n+8}{7n+15}$

$\Rightarrow \frac{a_1+\frac{(n-1)}{2}{d}_1}{a_2+\frac{(n-1)}{2}{d}_2}=\frac{3n+8}{7n+15}$[$\frac{1}{2}$]

Replacing $\frac{n-1}{2}$ by 11, i.e., n by 23 on both sides, we get

$\frac{a_1+\frac{(23-1)}{2}{d}_1}{a_2+\frac{(23-1)}{2}{d}_2}=\frac{3\times 23+8}{7\times 23+15}\Rightarrow \frac{a_1+11d_1}{a_2+11d_2}=\frac{77}{176}=\frac{7}{16}$

$\text{Hence, the required ratio is}7:16$