The number of values of a in N such that the | vedclass

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Question

Mean $=13$

Variance $=\frac{9+49+144+ a ^{2}+(43- a )^{2}}{5}-13^{2} \in N$

$\Rightarrow \frac{2 a^{2}-a+1}{5} \in N$

$\Rightarrow 2 a^{2}-a+

Vedclass · Accepted Answer

$0$

Vedclass · Answer

Mean $=13$

Variance $=\frac{9+49+144+ a ^{2}+(43- a )^{2}}{5}-13^{2} \in N$

$\Rightarrow \frac{2 a^{2}-a+1}{5} \in N$

$\Rightarrow 2 a^{2}-a+1-5 n=0$ must have solution as natural numbers

its $D=40 n-7$ always has $3$ at unit place

$\Rightarrow D$ can't be perfect square

So, a can't be integer.

The number of values of $a \in N$ such that the variance of $3,7,12 a, 43-a$ is a natural number is (Mean $=13$)

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