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Three numbers are in $AP$ such that their sum is $18$ and sum of their squares is $158$ . What is the greatest number among them?

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1

11

12

None of these

answer is B.

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Detailed Solution

Given, three numbers are in

AP

such that their sum is

18

and sum of their squares is

158

.
Let the three terms of the AP, be,

a - d, a, a + d

.
Then,

(a - d) + (a) + (a + d) = 18

\Rightarrow 3 a = 18

\Rightarrow a = 6

Therefore,

6 - d, 6, 6 + d

.
As given, the sum of square = 158.

\Rightarrow (6 - d)^{2} + 6^{2} + (6 + d)^{2} = 158

⇒

6^{2} + d^{2} - 12 d + 6^{2} + 6^{2} + d^{2} + 12 d = 158

⇒

108 + 2 d^{2} = 158

⇒

2 d^{2} = 50

⇒

d^{2} = 25

\Rightarrow d = \pm 5

When

d = + 5

,
The terms are 1, 6, 11.
When

d = - 5

,
The terms are 11, 6, 1.
Therefore, the greatest among all the terms is 11.
Hence, the correct option is 2.

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