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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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