Question 28 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.
We know that
Sn = n/2 ( 2a + (n – 1)d )
Where, Sn = sum of n terms of A.P.
n = number of terms
a = first term and d = common difference
Now,
Sum of first m terms = Sm
Sm = m/2 [2a + (m – 1)d]
Sum of first n terms = Sn
Sn = n/2 [2a + (n – 1)d]
It is given that
Sum of first m terms = Sum of first n terms
m/2 [2a + (m – 1)d] = n/2 [2a + (n – 1)d]
Cancelling denominator 2 both sides
m[2a + (m – 1)d] = n[2a + (n – 1)d]
2am + md(m – 1) = 2an + nd(n – 1)
2am – 2an = nd (n – 1) – md(m – 1)
2a(m – n) = d[n(n – 1) – m(m – 1)]
2a(m – n) = d[n2 – n – (m2 – m)]
2a(m – n) = d[n2 – n – m2 + m]
2a(m – n) = d[n2 – m2 + m – n]
2a(m – n) = – d [ –n2 + m2 – m + n]
2a(m – n) = – d [m2 – n2 – (m – n)]
2a(m – n) = – d [(m – n) (m + n) – (m – n)]
2a(m – n) = – d(m – n) [m + n – 1]
2a(m – n) + d(m – n) [m + n – 1] = 0
(m – n) [2a + d(m + n – 1)] = 0
Therefore,
m – n = 0
m = n
Not possible as m ≠ n
2a + d(m + n – 1) = 0
So,
2a + d(m + n – 1) = 0
Now, finding sum of first (m + n) terms
We know that,
Sum of n terms = n/2 [2a + (n – 1)d]
For sum of (m + n) terms, we mut n = (m + n)
Sum of (m + n) term is = (m + n)/2 [2a + (m + n – 1)d]
= (m + n)/2 × 0
= 0
∴ Sum of (m + n) term is 0
Hence proved

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.