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प्रश्न
The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.
उत्तर
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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