If ( m+1 ) th term of an ap is twice the ( n+1) thterm prove that (3m+1) th term is twice the ( m+n+1)th term .... please answer
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hey mate
here's the proof
here's the proof
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Step-by-step explanation:
Note : nth term of an AP is an = a + (n - 1) * d.
(i)
Given that (m + 1)th term of an AP is twice the (n + 1)th term.
⇒ a + (m + 1 - 1) * d = 2[a + (n + 1 - 1) * d]
⇒ a + md = 2[a + nd]
⇒ a + md = 2a + 2nd
⇒ -a = 2nd - md
⇒ a = md - 2nd
⇒ a = (m - 2n)d
(ii)
Consider, (3m + 1)th term:
⇒ a + (3m + 1 - 1) * d
⇒ a + 3md
⇒ (m - 2n) * d + 3md
⇒ md - 2nd + 3md
⇒ 4md - 2nd
⇒ (4m - 2n)d ------ (1)
(iii)
Consider, (m + n + 1)th term.
⇒ a + (m + n + 1 - 1) * d
⇒ a + md + nd
⇒ (m - 2n) * d + md + nd
⇒ md - 2nd + md + nd
⇒ 2md - nd
⇒ (2m - n)d ---- (2)
Therefore, From (1) & (2),
We can say that (3m + 1)Th term is twice the (m + n + 1)th term.
Hope it helps!
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