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Question
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
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Solution
Let the required terms be (a-d) , a and ( a+d)
Then (a-d) + a+ (a+d) = 21
⇒ 3a = 21
⇒ a =7
Also , `(a - d)^2 + a^2 + (a + d)^2 = 165`
⇒ `3a^2 + 2d^2 = 165 `
⇒ `(3 xx 49 +2d^2) = 165`
⇒`2d^2 = 165 - 147 = 18`
⇒`d^2 = 9`
⇒ `d = +- 3`
Thus ,`a = 7 and d = +- 3`
Hence, the required terms are (4,7,10) or (10,7,4).
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