Let a_1, a_2, ldots, a_n be n non-zero real n | vedclass

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(C) Let $p$ be the number of positive real numbers and $(n-p)$ be the number of negative real numbers.The product $a_j a_k$ is positive if both $a_j$

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(C) Let $p$ be the number of positive real numbers and $(n-p)$ be the number of negative real numbers.The product $a_j a_k$ is positive if both $a_j$ and $a_k$ have the same sign (both positive or both negative).Thus,the number of pairs with a positive product is $\binom{p}{2} + \binom{n-p}{2} = 55$.The product $a_j a_k$ is negative if one is positive and the other is negative.Thus,the number of pairs with a negative product is $\binom{p}{1} 	imes \binom{n-p}{1} = p(n-p) = 50$.Expanding the first equation:$\frac{p(p-1)}{2} + \frac{(n-p)(n-p-1)}{2} = 55$$p^2 - p + (n-p)^2 - (n-p) = 110$$p^2 + (n-p)^2 - (p + n - p) = 110$$p^2 + (n-p)^2 - n = 110$We know that $(p + (n-p))^2 = p^2 + (n-p)^2 + 2p(n-p) = n^2$.So,$p^2 + (n-p)^2 = n^2 - 2p(n-p) = n^2 - 2(50) = n^2 - 100$.Substituting this into the expanded equation:$(n^2 - 100) - n = 110$$n^2 - n - 210 = 0$$(n - 15)(n + 14) = 0$.Since $n > 0$,we have $n = 15$.Now,$p(15 - p) = 50$ $\Rightarrow p^2 - 15p + 50 = 0$ $\Rightarrow (p - 10)(p - 5) = 0$.Thus,$p = 5$ or $p = 10$.In either case,$p^2 + (n-p)^2 = 5^2 + 10^2 = 25 + 100 = 125$.

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