Question

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 1

Answer:

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Answer 2

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!