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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
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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