| | Question 35

A number 21 is divided into three parts which are in AP and sum of their squares is 155. Find the largest number.
(A) 4
(B) 7
(C) 9
(D) 11


Answer: (C)

Explanation: Let the three consecutive parts of AP are (a-d), a, (a+d).
Given that
(a-d) + a + (a+d) = 21
3a = 21
a = 7
Again, (a-d)2 + a2 + (a+d)2 = 155
a2 + d2 – 2ad + a2 + a2 + d2 + 2ad = 155
3a2 + 2d2 = 155
put value of a
3(7)2 + 2d2 = 155
2d2 = 155 – 147
d2 = 4
d = ∓2
Hence, the largest part is (a+d) = 7+2 = 9


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