Question

The sum of the 2nd term and the 7th term of an A.P. is 30. If its 15th term is 1 less than twice its 8th term, find the A.P.

Answer 1

Let a be the first term and d be the common difference of the A.P.

Then, a₂ + a₇ = 30   ...(Given)

∴ (a + d) + (a + 6d) = 30   ...[a_n = a + (n – 1)d]

`\implies` 2a + 7d = 30  ...(1)

Also,

a₁₅ = 2a₈ – 1   ...(Given)

`\implies` a + 14d = 2(a + 7d) – 1

`\implies` a + 14d = 2a + 14d – 1

`\implies` –a = –1

`\implies` a = 1

Putting a = 1 in (1), we get

2 × 1 + 7d = 30

`\implies` 7d = 30 – 2 = 28

`\implies` d = 4

So,

a₂ = a + d = 1 + 4 = 5

a₃ = a + 2d = 1 + 2 × 4 = 9 ....... 

Hence, the A.P. is 1, 5, 9, 13, .......