Question

Q18 If in an AP, the sum of m terms is equal to n and the sum of n terms is equal
to m, then prove that the sum of (m+n) terms is - (m+n).​

Answer 1

Given :

In an AP, the sum of m terms, (Sm) = n.

The sum of n terms, (Sn) = m.

To prove :

The sum of (m+n) term is - (m+n).

Proof :

Let ‘a’ be the first term and d is the common difference in given AP.

So, $\sf S_m = S_n$

Where,

• $\sf S_m = Sum \ of \ m \ terms.$

• $\sf S_n = Sum \ of \ n \ terms.$

Now,

$\implies \sf \dfrac {m}{2} [2a + (m-1)d] = n$

$\implies \sf 2am + m(m-1)d = 2n \ --(i)$

Also,

$\sf For S_n = m,$

$\implies \sf \dfrac {n}{2} [2a + (n-1)d] = 2m$

$\implies \sf 2an + n(n-1)d = 2m \ -- (ii)$

Here, Subtracting equation (ii) from (i),

$\implies \sf 2a(m-n) + [m(m-1) - n(n-1)d] = 2n - 2m$

$\implies \sf 2a(m-n) + [(m^2 - n^2) - (m-n)d] = -2(m-n)$

$\implies \sf 2a + (m+n-1)d = -2 \ -- (iii)$

Divide the both sides by (m-n).

We get,

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

$\implies \sf S_{m+n} = \dfrac {m+n}{2} (-2)$

$\implies \sf S_{m+n} = -(m+n)$

∴ Hence proved.

Answer 2

Step-by-step explanation:

Given:-

● If in an AP, the sum of m terms is equal to n and the sum of n terms is equal to m.

To Prove:-

● The sum of (m + n) terms is - (m + n).

Solution:-

☯️ Let a be the first term and d be the common difference of the given A.P. Then,

$\bf : \implies {\underline{S_{m}\ =\ n}}$

$\sf : \implies {\dfrac{m}{2}\ [2a\ +\ (m\ -\ 1)d]\ =\ n}$

$\sf : \implies {2am\ +\ m(m\ -\ 1)d\ =\ 2n\ ..........(1).}$

$\bf : \implies {\underline{S_{n}\ =\ m}}$

$\sf : \implies {\dfrac{n}{2}\ [2a\ +\ (n\ -\ 1)d]}$

$\sf : \implies {2an\ +\ n(n\ -\ 1)d\ =\ 2m\ ..........(2).}$

Subtracting equation (2) from equation (1), we get:-

$\sf : \implies {2a(m\ -\ n)\ +\ [m(m\ -\ 1)\ -\ n(n\ -\ 1)]d\ =\ 2n\ -\ 2m}$

$\sf : \implies {2a(m\ -\ n)\ +\ [(m²\ -\ n²)\ -\ (m\ -\ n)]d\ =\ -2(m\ -\ n)}$

$\sf : \implies {2a\ +\ (m\ +\ n\ -\ 1)d\ =\ -2\ ......[On\ dividing\ both\ sides\ by\ (m\ -\ n)]\ ...........(3).}$

Now here:-

$\sf : \implies {S_{m+n}\ =\ \dfrac{m+n}{2}\ [2a\ +\ (m\ +\ n\ -\ 1)d]}$

$\sf : \implies {S_{m+n}\ =\ \dfrac{m+n}{2}\ (-2)\ ........[Using\ (3)]}$

$\sf : \implies {S_{m+n}\ =\ -\ (m\ +\ n)}$

Hence proved ✔.