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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 of its 8th term, find the A.P.
Solution
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 ...[an = 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, .......
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