Mathematics
The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.
Answer
Let three consecutive terms be (a - d), a, (a + d)
S = 21
⇒ a - d + a + a + d = 21
⇒ 3a = 21
⇒ a = 7.
Given, sum of their squares is 165
⇒ (a - d)2 + a2 + (a + d)2 = 165
⇒ (7 - d)2 + 72 + (7 + d)2 = 165
⇒ 49 + d2 - 14d + 49 + 49 + d2 + 14d = 165
⇒ 147 + 2d2 = 165
⇒ 2d2 = 165 - 47 = 18
⇒ d2 = 9
⇒ d = ± 3
Let d = 3,
A.P. = (7 - 3), 7, (7 + 3) = 4, 7, 10.
Let d = -3,
A.P. = (7 - (-3)), 7, (7 + (-3)) = 10, 7, 4.
Hence, the three consecutive terms of A.P. = 4, 7, 10 or 10, 7, 4.
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