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

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Question

Let the A.P is

$a-2 d, a-d, a, a+d, a+2 d$

$\because \operatorname{sum}=25 \Rightarrow \mathrm{a}=5$

Product $=2520$

$\left(25-4 d^{2}\rig

Vedclass · Accepted Answer

$16$

Vedclass · Answer

Let the A.P is

$a-2 d, a-d, a, a+d, a+2 d$

$\because \operatorname{sum}=25 \Rightarrow \mathrm{a}=5$

Product $=2520$

$\left(25-4 d^{2}\right)\left(25-d^{2}\right)=504$

$4 \mathrm{d}^{4}-125 \mathrm{d}^{2}+121=0$

$\Rightarrow \mathrm{d}^{2}=1, \frac{121}{4}$

$\Rightarrow \mathrm{d}=\pm 1, \pm \frac{11}{2}$

$\mathrm{d}=\pm 1$ is rejected because none of the term can be $\frac{-1}{2}$

$\Rightarrow \mathrm{d}=\pm \frac{11}{2}$

$\Rightarrow$ AP will be $-6,-\frac{1}{2}, 5, \frac{21}{2}, 16$

Largest term 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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