The sum of first q terms of an A.P. is 162. The ratio of its 6th term

1.	The sum of first q terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1: 2. Find the first and 15th term of the A.P.
Answer» Let a be the first term and d be the common difference. And we know that, sum of first n terms is: S_n = \(\frac{n}{2}\)(2a + (n − 1)d) Also, nth term is given by a_n = a + (n – 1)d From the question, we have S_q = 162 and a₆: a₁₃ = 1 : 2 So, 2a₆= a₁₃ ⟹ 2 [a + (6 – 1d)] = a + (13 – 1)d ⟹ 2a + 10d = a + 12d ⟹ a = 2d …. (1) And, S₉ = 162 ⟹ S₉ = \(\frac{9}{2}\)(2a + (9 − 1)d) ⟹ 162 = \(\frac{9}{2}\)(2a + 8d) ⟹ 162 × 2 = 9[4d + 8d] [from (1)] ⟹ 324 = 9 × 12d ⟹ d = 3 ⟹ a = 2(3) [from (1)] ⟹ a = 6 Hence, the first term of the A.P. is 6 For the 15^th term, a₁₅ = a + 14d = 6 + 14 × 3 = 6 + 42 Therefore, a₁₅ = 48

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

Answer»

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

And we know that, sum of first n terms is:

S_n = \(\frac{n}{2}\)(2a + (n − 1)d)

Also, nth term is given by a_n = a + (n – 1)d

From the question, we have

S_q = 162 and a₆: a₁₃ = 1 : 2

So,

2a₆= a₁₃

⟹ 2 [a + (6 – 1d)] = a + (13 – 1)d

⟹ 2a + 10d = a + 12d

⟹ a = 2d …. (1)

And, S₉ = 162

⟹ S₉ = \(\frac{9}{2}\)(2a + (9 − 1)d)

⟹ 162 = \(\frac{9}{2}\)(2a + 8d)

⟹ 162 × 2 = 9[4d + 8d] [from (1)]

⟹ 324 = 9 × 12d

⟹ d = 3

⟹ a = 2(3) [from (1)]

⟹ a = 6

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

For the 15^th term, a₁₅ = a + 14d = 6 + 14 × 3 = 6 + 42

Therefore, a₁₅ = 48