If the sum of first m terms of an Ap is same as the sum of its first n terms (m≠n) show that the sum of it's first (m+n) term is zero
Answers
Answer:
Step-by-step explanation:
Let a be the first term and d is common difference of the A.P then sum of n terms in A.P
is Sn = (n/2)[ 2a + (n - 1) d]
Similarly um of m terms in A.P is Sm = (m/2)[ 2a + (m - 1) d]
Given that sum of first m terms of an AP is the same as the sum of its first n terms i.e
Sm = Sn
⇒ (m/2)[ 2a + (m - 1) d] = (n/2)[ 2a + (n - 1) d]
⇒ (m/2)[ 2a + (m - 1) d] - (n/2)[ 2a + (n - 1) d] = 0
⇒ 2a[ (m - n)/2)] +[ (1/2)(m2- m - n2 + n) d] = 0
⇒ 2a[ (m - n)/2)] +[ (1/2)((m2 - n2) - (m - n)) d] = 0
⇒ 2a[ (m - n)/2)] +[ (1/2)((m - n)(m + n) - (m - n)) d] = 0
⇒ (m - n)/2 [2a +[ (m + n) - 1) d] = 0
⇒ [2a +[ (m + n) - 1) d] = 0
∴ 2a = - [ (m + n) - 1) d -----------(1)
Sum of first ( m + n ) terms
Sm +n = (m+n) / 2 [ 2a + ( m+ n -1) d]
Sm +n = (m+n) / 2 [ - [ (m + n) - 1) d + ( m+ n -1) d] [ from (1) ]
Sm +n = 0.