Question

Three number are in A.P. such that their sum is 18 and sum of their squares is 158. The greatest number among them is.

• A. 10
• B. 11
• C. 12
• D. None of these

Answer 1

Answer: (B)

11

Answer 2

Let three number of A.P. $a - d , a$ $a + d$

Sum = 18, and $( a - d ) ^ { 2 } + a ^ { 2 } + ( a + d ) ^ { 2 } = 58$

$a - d + a + a + d = 18$

$a = 6$ and $( 6 - d ) ^ { 2 } + 36 + ( 6 + d ) ^ { 2 } = 158$

= $36 + d ^ { 2 } + 36 + d ^ { 2 } = 122$ $= 2 d ^ { 2 } + 72 = 122$

$= 2 d ^ { 2 } = 50$ ⇒ $d = 5$ .

Hence Numbers are 1, 6, 11, i.e. maximum number is 11.