Question

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

Answer 1

Let a be the first term and d be the common difference.
We know that, n^th term = a_n = 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 = \[\frac{3}{2} \times 2\]  [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 5, 7, 9, .......