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The sum of 5 consecutive terms of an arithmetic series is 30 and the sum of the squares of these terms is 220. Find the terms.

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As the number of terms is odd, let the middle $\left(\frac{5+1}2=3\text{rd}\right)$ term be $a$

So, the $5$ terms will be $a,a\pm d,a\pm 2d$

So, $(a-2d)+(a-d)+a+(a+d)+(a+2d)=30\implies 5a=30\implies a=6$

and $(a-2d)^2+(a-d)^2+a^2+(a+d)^2+(a+2d)^2=220$

$\implies a^2+2\{a^2+d^2\}+2\{a^2+(2d)^2\}=220$ as $(A+B)^2+(A-B)^2=2(A^2+B^2)$

$\implies 10d^2=220-5\cdot 6^2=40\implies d^2=4$

Had the number of terms been even, we could take the terms as $a\pm d,a\pm3d,\cdots$

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