The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165.

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)+ a2 + (a + d)2​ = 165
⇒ 3a2 + 2d2 = 165  

$\Rightarrow\left(3 \times 49+2 d^{2}\right)=165$

$\Rightarrow 2 d^{2}=165-147=18$

$\Rightarrow d^{2}=9$

$\Rightarrow d=\pm 3$

Thus, $a=7$ and $d=\pm 3$

 Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4).

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