Question:
If the sum of first m terms of an AP is same as the sum of its first n terms show that the sum of its first m n terms is zero
Answer:
Let a is the first term and d is the common difference of AP.
We know sum of n terms of AP = (n/2)*{2a + (n - 1)d}
Therefore sum of m terms = (m/2)*{2a + (m - 1)d} ............1
sum of n terms = (n/2)*{2a + (n - 1)d} ..................2
Given sum of m terms of A.P. is the same as the sum of its n terms
=> (m/2)*{2a + (m - 1)d} = (n/2)*{2a + (n - 1)d}
=> 2ma + (m2 -m)d = 2na + (n2 - n)d
=> (2m - 2n)a = (n2 - n)d - (m2 - m)d
=> 2(m - n)a = n2 d - nd - m2 d + md
=> 2a(m - n) = -d(m -n)*(m + n - 1)
=> a = -d/2(m + n - 1) ..................3
Now sum of m + n terms = {(m + n)/2}*[-2d/{2(m + n - 1)} + (m + n - 1)d]
= {(m + n)/2}*[-d(m + n - 1)} + (m + n - 1)d]
= 0
Hence, the sum of its (m + n) terms is zero.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
