The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.
Let the numbers be (a - d), a, (a + d).
Now, sum of the numbers = 21
∴ (a - d) + a + (a + d) = 21
⇒ 3a = 21
⇒ a = 7
Now, sum of the squares of the terms = 165
⇒ (a - d)2 + a2 + (a + d)2 = 165
⇒ a2 + d2 - 2ad + a2 + a2 + d2 + 2ad = 165
⇒ 3a2 + 2d2 + a = 165
Put the value of a = 7, we get,
3(49) + 2d2 = 165
⇒ 2d2 = 165 - 147
⇒ 2d2 = 18
⇒ d2 = 9
⇒ d = 
3
∴ If d = 3, then the numbers are 4, 7, 10.
If d = - 3, then the numbers are 10, 7, 4.
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