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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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