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

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8. Sequences and Series

easy

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

Solution

(b) 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 $11$.

Standard 11

Mathematics

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