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  • Let <img alt="\mathrm{a_1, a_2, a_3, \ldots be\: a \: G P}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba_1%2C%20a_2%2C%20a_3%2C%20%5Cldots%20be%5C%3A%20a%20%5C%3A%20G%20P%7D"

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

Option: 1

60


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let first term of G.P be a with common ratio \mathrm{r}
\begin{aligned} & \text { Given : } a_4 \cdot a_6=9 \\ \end{aligned}

                \begin{aligned} & a_5+a_7=24 \\ & a_4=a^3, a_5=a^4, a_6=a r^5, a_7=a r^6 \\ & a_4 \cdot a_6=a^2 r^8=9 \end{aligned}

\begin{aligned} & \Rightarrow \quad a^4=3 \\ \end{aligned}

          \begin{aligned} \quad a_5=3 \\ \end{aligned}

\begin{aligned} & \therefore \quad a_7=24-3=21 \\ & \Rightarrow \quad \frac{a_7}{a_5}=r^2=7 \\ & \Rightarrow \quad r=\sqrt{7}, a=\frac{3}{49} \\ & a_1 a_9+a_2 a_4 a_9+\quad a_5+a_7=a_1 a_9+\left(a_1\right)\left(a^3\right) a_9+24 \\ & =a_1 a_9+a_1\left(a^4\right) a_9+24 \\ & =a_1 a_9\left(1+a_5\right)+24=\left(a^4\right)^2(4)+24 \\ & =36+24=60 \end{aligned}

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Divya Prakash Singh

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