Three number are in A.P. such that their sum | vedclass

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Question

(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

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(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 $11$.

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

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