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Q.

If the integer k is added to each of the numbers 36,300,596; squares of three consecutive terms of an A.P are obtained. Then the last digit of k is

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answer is 5.

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Detailed Solution

Let the A. P. be a, a + d, a + 2d
a2=36+k,(a+d)2=300+k,(a+2d)2=596+k⇒ 2ad+d2=300−36=264 and 2ad+3d2=596−300=296d2=16,d=4,a=31∴ k=a2−36=312−36=925

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