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Question
The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.
Solution
Let the three consecutive terms in A.P. be a – d, a and a + d.
∴ (a – d) + a + (a + d) = 21
`=>` a = 7 ...(1)
Also, (a – d)2 + a2 + (a + d)2 = 165
`=>` a2 + d2 - 2ad + a2 + a2 + d2 + 2ad = 165
`=>` 3a2 + 2d2 = 165
`=>` 3 × (7)2 + 2d2 = 165 ...[From (1)]
`=>` 3 × 49 + 2d2 = 165
`=>` 147 + 2d2 = 165
`=>` 2d2 = 18
`=>` d2 = 9
`=>` d = ±3
When a = 7 and d = 3
Required terms = a – d, a and a + d
= 7 – 3, 7, 7 + 3
= 4, 7, 10
When a = 7 and d = –3
Required terms = a – d, a and a + d
= 7 – (–3), 7, 7 + (–3)
= 10, 7, 4
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