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 first (m + n) terms is zero.​

Answer 1

ATQ,

Sum of 'm' terms of an AP = Sum of 'n' terms of the same AP.

$\Longrightarrow$ S$\sf _m$ = S$\sf _n$

$\sf \Longrightarrow \dfrac{m}{2} \left(2a+(m-1)d\right) = \dfrac{n}{2} \left(2a+(n-1)d\right)$

Cancelling 2 we get,

$\sf \Longrightarrow m \left(2a+(m-1)d\right) = n \left(2a+(n-1)d\right)$

$\sf \Longrightarrow m \left(2a+dm-d\right) = n \left(2a+dn-d\right)$

$\sf \Longrightarrow 2am+dm^2-dm = 2an+dn^2-dn$

$\sf \Longrightarrow 2am+dm^2-dm-2an-dn^2+dn = 0$

Re-arranging according to the like terms we get,

$\sf \Longrightarrow 2am-2an+dm^{2}-dn^2-dm+dn = 0$

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

Using a² - b² = (a - b)(a + b) we get,

$\sf \Longrightarrow 2a(m-n)+d(m-n)(m+n)-d(m-n) = 0$

Taking 'd' out as a common factor we get,

$\sf \Longrightarrow 2a(m-n)+d\left[(m-n)(m+n)-(m-n)\right] = 0$

Taking (m - n) out as a common factor we get,

$\sf \Longrightarrow (m - n) (2a+d(m+n-1) = 0$

Taking (m - n) to the other side we get,

$\sf {\Longrightarrow \left(2a+d \ (m+n-1) = 0} \longrightarrow \sf \textcircled{\scriptsize1}$

Now, We find the sum of the (m + n)th terms.

$\Longrightarrow \sf {S_n=\dfrac{n}{a}(2a+(n-1)d)}$

Here, n = (m + n)

$\Longrightarrow \sf S_{(m+n)} = \dfrac{m+n}{a}(2a+((m+n) - 1)d)$

Substituting Equation 1 above we get,

$\Longrightarrow \sf S_{(m+n)} = \dfrac{m+n}{a} \times 0$

$\Longrightarrow \sf S_{(m+n)} = 0$

Hence Proved.