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Question

# The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.

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Solution

## Let a be the first term and d be the common difference. We know that, nth term = an = a + (n − 1)d According to the question, a19 = 3a6 ⇒ a + (19 − 1)d = 3(a + (6 − 1)d) ⇒ a + 18d = 3a + 15d ⇒ 18d − 15d = 3a − a ⇒ 3d = 2a ⇒ a = $\frac{3}{2}$d .... (1) Also, a9 = 19 ⇒ a + (9 − 1)d = 19 ⇒ a + 8d = 19 ....(2) On substituting the values of (1) in (2), we get $\frac{3}{2}$d + 8d = 19 ⇒ 3d + 16d = 19 × 2 ⇒ 19d = 38 ⇒ d = 2 ⇒ a = $\frac{3}{2}×2$ [From (1)] ⇒ a = 3 Thus, the A.P. is 3, 5, 7, 9, .... .

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