

InterviewSolution
1. |
If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term. |
Answer» Given : (m + 1)th term of an A.P. is twice the (n + 1)th term ⇒ am+1 = 2an+1 To prove : a3m+1 = 2am+n+1 We know, an = a + (n – 1)d Where a is first term or a1 and d is common difference and n is any natural number When n = m + 1 : ∴ am+1 = a + (m + 1 – 1)d ⇒ am+1 = a + md When n = n + 1 : ∴ an+1 = a + (n + 1 – 1)d ⇒ an+1 = a + nd According to question : am+1 = 2an+1 ⇒ a + md = 2(a + nd) ⇒ a + md = 2a + 2nd ⇒ a – 2a + md – 2nd = 0 ⇒ -a + (m – 2n)d = 0 ⇒ a = (m – 2n)d………(i) an = a + (n – 1)d When n = m + n + 1 : ∴ am+n+1 = a + (m + n + 1 – 1)d ⇒ am+n+1 = a + md + nd ⇒ am+n+1 = (m – 2n)d + md + nd……… (From (i)) ⇒ am+n+1 = md – 2nd + md + nd ⇒ am+n+1 = 2md – nd………(ii) When n = 3m + 1 : ∴ a3m+1 = a + (3m + 1 – 1)d ⇒ a3m+1 = a + 3md ⇒ a3m+1 = (m – 2n)d + 3md……… (From (i)) ⇒ a3m+1 = md – 2nd + 3md ⇒ a3m+1 = 4md – 2nd ⇒ a3m+1 = 2(2md – nd) ⇒ a3m+1 = 2am+n+1…………(From (ii)) Hence Proved. |
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