Let a_1, a_2, a_3, ldots be a GP of increasin | vedclass

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(C) Let the $GP$ be $a, ar, ar^2, \ldots$. Given $a_4 \cdot a_6 = 9$,we have $(ar^3)(ar^5) = 9$,which implies $a^2 r^8 = 9$,so $a_5^2 = 9$. Since the

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$60$

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(C) Let the $GP$ be $a, ar, ar^2, \ldots$. Given $a_4 \cdot a_6 = 9$,we have $(ar^3)(ar^5) = 9$,which implies $a^2 r^8 = 9$,so $a_5^2 = 9$. Since the terms are positive,$a_5 = 3$. Given $a_5 + a_7 = 24$,we have $a_5 + a_5 r^2 = 24$. Substituting $a_5 = 3$,we get $3(1 + r^2) = 24$,so $1 + r^2 = 8$,which gives $r^2 = 7$. Since $a_5 = ar^4 = 3$,we have $a(7^2) = 3$,so $a = \frac{3}{49}$. Now,calculate the expression $a_1 a_9 + a_2 a_4 a_9 + a_5 + a_7$: $a_1 a_9 = a(ar^8) = a^2 r^8 = (ar^4)^2 = a_5^2 = 3^2 = 9$. $a_2 a_4 a_9 = (ar)(ar^3)(ar^8) = a^3 r^{12} = (ar^4)^3 = a_5^3 = 3^3 = 27$. $a_5 + a_7 = 24$. Thus,$9 + 27 + 24 = 60$.

Let $a_1, a_2, a_3, \ldots$ be a $GP$ of increasing positive numbers. If the product of the fourth and sixth terms is $9$ and the sum of the fifth and seventh terms is $24$,then $a_1 a_9 + a_2 a_4 a_9 + a_5 + a_7$ is equal to $.........$.

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