Five numbers are in A.P.,whose sum is 25 and | vedclass

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(C) Let the five numbers in $A.P.$ be $(a-2d, a-d, a, a+d, a+2d)$.Sum $= (a-2d) + (a-d) + a + (a+d) + (a+2d) = 5a = 25$,so $a = 5$.The numbers are $(5

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Let the five numbers in $A.P.$ be $(a-2d, a-d, a, a+d, a+2d)$.Sum $= (a-2d) + (a-d) + a + (a+d) + (a+2d) = 5a = 25$,so $a = 5$.The numbers are $(5-2d, 5-d, 5, 5+d, 5+2d)$.The product is $(5-2d)(5+2d)(5-d)(5+d)(5) = 2520$.$(25-4d^2)(25-d^2) = 504$.$625 - 25d^2 - 100d^2 + 4d^4 = 504$.$4d^4 - 125d^2 + 121 = 0$.Let $x = d^2$,then $4x^2 - 125x + 121 = 0$.$(4x - 121)(x - 1) = 0$,so $d^2 = 1$ or $d^2 = \frac{121}{4}$.If $d^2 = 1$,$d = \pm 1$,the terms are $(3, 4, 5, 6, 7)$ or $(7, 6, 5, 4, 3)$,none of which is $-\frac{1}{2}$.If $d^2 = \frac{121}{4}$,$d = \pm \frac{11}{2}$.For $d = \frac{11}{2}$,the terms are $(5-11, 5-5.5, 5, 5+5.5, 5+11) = (-6, -0.5, 5, 10.5, 16)$.Since $-\frac{1}{2}$ is present,this is the correct sequence.The greatest number is $16$.

Five numbers are in $A.P.$,whose sum is $25$ and product is $2520$. If one of these five numbers is $-\frac{1}{2}$,then the greatest number amongst them is

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