Question

The sum of first 9 terms of an A.P. is 162. The ratio of its 6^th term to its 13^th term is 1 : 2. Find the first and 15^th term of the A.P.

Answer 1

Let a be the first term and d be the common difference.

We know that, sum of first n terms = S_n = \[\frac{n}{2}\][2a + (n − 1)d]

Also, n^th term = a_n = a + (n − 1)d

According to the question,

S_q = 162 and \[\frac{a_6}{a_{13}} = \frac{1}{2}\]

Now,

\[\frac{a_6}{a_{13}} = \frac{1}{2}\]

\[ \Rightarrow \frac{a + (6 - 1)d}{a + (13 - 1)d} = \frac{1}{2}\]

\[ \Rightarrow \frac{a + 5d}{a + 12d} = \frac{1}{2}\]

\[ \Rightarrow 2a + 10d = a + 12d\]

\[ \Rightarrow 2a - a = 12d - 10d\]

\[ \Rightarrow a = 2d . . . . . (1)\]

Also,
S₉ =\[\frac{9}{2}\][2a + (9 − 1)d]

⇒ 162 = \[\frac{9}{2}\][2(2d) + 8d]           [From (1)]

⇒ 18 = \[\frac{1}{2}\] ⇒ 18 = 6d

⇒ d = 3

⇒ a = 2 × 3                              [From (1)]

⇒ a = 6

Thus, the first term of the A.P. is 6.

Now,
a₁₅ = 6 + (15 − 1)3

= 6 + 42

= 48
Thus, 15^th term of the A.P. is 48.