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.


Disclaimer: There is a misprint in the question, 'the sum of 9 terms' should be witten instead of 'the sum of q terms'.

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,
$$\begin{gathered} \frac{a_6}{a_{13}}=\frac{1}{2} \\ \Rightarrow \frac{a+(6-1)d}{a+(13-1)d}=\frac{1}{2} \\ \Rightarrow \frac{a+5d}{a+12d}=\frac{1}{2} \\ \Rightarrow 2a+10d=a+12d \\ \Rightarrow 2a-a=12d-10d \\ \Rightarrow a=2d.....\left(1\right) \end{gathered}$$

Also,
S₉ = $\frac{9}{2}$[2a + (9 − 1)d]
⇒ 162 = $\frac{9}{2}$[2(2d) + 8d] [From (1)]
⇒ 18 = $\frac{1}{2}$[12d]
⇒ 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.