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 $(a-2d, a-d, a, a+d, a+2d)$.Given the sum is $25$,so $(a-2d) + (a-d) + a + (a+d) + (a+2d) = 25$,which simplifies

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$16$

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(C) Let the five numbers in $A.P.$ be $(a-2d, a-d, a, a+d, a+2d)$.Given the sum is $25$,so $(a-2d) + (a-d) + a + (a+d) + (a+2d) = 25$,which simplifies to $5a = 25$,so $a = 5$.The product is $(a-2d)(a-d)(a)(a+d)(a+2d) = 2520$.Substituting $a=5$,we get $5(25-4d^2)(25-d^2) = 2520$,which simplifies to $(25-4d^2)(25-d^2) = 504$.Expanding this,we get $625 - 25d^2 - 100d^2 + 4d^4 = 504$,or $4d^4 - 125d^2 + 121 = 0$.Factoring the quadratic in $d^2$,we get $(4d^2 - 121)(d^2 - 1) = 0$,so $d^2 = 1$ or $d^2 = \frac{121}{4}$.If $d^2 = 1$,the terms are $3, 4, 5, 6, 7$ or $7, 6, 5, 4, 3$. None of these is $-\frac{1}{2}$.If $d^2 = \frac{121}{4}$,then $d = \pm \frac{11}{2}$.For $d = \frac{11}{2}$,the terms are $5-11, 5-5.5, 5, 5+5.5, 5+11$,which are $-6, -0.5, 5, 10.5, 16$.Since one of the numbers is $-\frac{1}{2}$,this sequence is valid. 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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