Question

# There are four numbers in A.P.  Their sum is 32 and sum of squares is 276. Find smallest of those numbers?

Solution

## Let us take the four numbers in A.P. to be $$a - 3d, a - d, a + d, a + 3d.$$Thus, their sum = $$4a = 32$$$$\Rightarrow a = 8.$$The sum of their squares is $$2(a^2 + d^2) + 2(a^2 + 9d^2) = 4a^2 + 20d^2 = 276$$$$\Rightarrow a^2 + 5d^2 = 69$$$$\Rightarrow 5d^2 = 69 - 64 = 5$$, implying d to be 1.The numbers are thus $$5, 7, 9,11$$. Mathematics

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