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

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(B) Let the three numbers in $A.P.$ be $(a - d)$,$a$,and $(a + d)$.Given that their sum is $18$:$(a - d) + a + (a + d) = 18$$3a = 18 \implies a = 6$.G

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Let the three numbers in $A.P.$ be $(a - d)$,$a$,and $(a + d)$.Given that their sum is $18$:$(a - d) + a + (a + d) = 18$$3a = 18 \implies a = 6$.Given that the sum of their squares is $158$:$(a - d)^2 + a^2 + (a + d)^2 = 158$$(6 - d)^2 + 6^2 + (6 + d)^2 = 158$$(36 - 12d + d^2) + 36 + (36 + 12d + d^2) = 158$$108 + 2d^2 = 158$$2d^2 = 158 - 108$$2d^2 = 50$$d^2 = 25 \implies d = \pm 5$.If $d = 5$,the numbers are $(6 - 5), 6, (6 + 5)$,which are $1, 6, 11$.If $d = -5$,the numbers are $(6 - (-5)), 6, (6 + (-5))$,which are $11, 6, 1$.In both cases,the greatest number is $11$.

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

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