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Question
How many terms are there in an A.P. whose first and fifth terms are – 14 and 2, respectively and the last term is 62?
Solution
a = –14, a5 = 2, an= 62
a5 = a + (5 – 1)d ...[∵ an = a + (n – 1)d]
`\implies` 2 = –14 + 4d
`\implies` 4d = 16
`\implies` d = 4
Now, an = 62
`\implies` a + (n – 1)d = 62
`\implies` –14 + (n – 1)4 = 62 ...(∵ a = –14, d = 4)
`\implies` (n – 1)4 = 76
`\implies` n – 1 = `76/4` = 19
`\implies` n = 19 + 1 = 20
∴ There are 20 terms in an A.P.
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