Question

If the sum of first m terms is equal to the sum of first n terms than show that the sum of first m+n terms is 0?

Answer 1

Solution: Given that Sₘ = Sₙ

→ m/2 × [2a + (m - 1)d] = n/2 × [2a + (n - 1)d]

→ m[2a + dm - d] = n[2a + dn - d]

→ 2am + dm² - dm = 2an + dn² - dn

→ 2am + dm² - dm - 2an - dn² + dn = 0

(Rearranging)

→ 2am - 2an - dm + dn + dm² - dn² = 0

(Taking out common terms)

→ 2a(m - n) - d(m - n) + d(m² - n²) = 0

→ 2a(m - n) - d(m - n) + d(m - n)(m + n) = 0

→ 2a(m - n) - d(m - n) + (m - n)(dm + dn) = 0

(Taking out common terms)

→ (m - n)[2a - d + dm + dn] = 0

→ [2a + (m + n - 1)d] = 0

→ Sₘ ₊ ₙ = (m + n)/2 × [2a + (m + n - 1)d]

→ Sₘ ₊ ₙ = (m + n)/2 × 0

→ Sₘ ₊ ₙ = 0

Hence Proved

Answer 2

Solution:

Let "a" be the first term and "d" be the common difference.

$\implies S_{m}=S_{n}\\ \\ \implies \dfrac{m}{2}[2a+(m-1)d]=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \implies 2am + (m-1)md=2an+(n-1)nd\\ \\ \implies 2a(m-n)=d[n^{2}-m^{2}-n+m]\\ \\ \implies 2a(m-n)=d(m-n)(1-m-n)\\ \\ \implies 2a=d(1-m-n)\;\;\;\;.............(1)$

Now,

$\implies S_{m+n}=\dfrac{(m+n)}{2}[2a+(m+n-1)d]\\ \\ \implies S_{m+n}=\dfrac{m+n}{2}[d(1-m-n)+(m+n-1)d]\;\;\;\;\;[By\;using\;eq\;(1)]\\ \\ \implies S_{m+n}=\dfrac{m+n}{2}[d-md-nd+md+nd-d]\\ \\ \implies S_{m+n}=0$

Hence Proved!!!