JEE Main 2023 — Mathematics Algebra

Given,

${a}_{1},{a}_{2},{a}_{3},\ldots$a1​,a2​,a3​,…. be a G.P. of increasing positive numbers,

And the sum of its ${6}^{\text{th }}$6th  and ${8}^{\text{th }}$8th  terms be $2$2

So, ${a}_{6}+{a}_{8}=2$a6​+a8​=2

$$\Rightarrow a{r}^{5}+a{r}^{7}=2.....(1)$$⇒ar5+ar7=2.....(1)

And the product of its ${3}^{\text{rd }}$3rd  and ${5}^{\text{th }}$5th  terms be $\frac{1}{9}$91​,

So, ${a}_{3}\cdot {a}_{5}=\frac{1}{9}\Rightarrow {a}^{2}\cdot {r}^{2}\cdot {r}^{4}=\frac{1}{9}$a3​⋅a5​=91​⇒a2⋅r2⋅r4=91​

$$\Rightarrow a{r}^{3}=\frac{1}{3}$$⇒ar3=31​

Now from putting the value of $a{r}^{3}=\frac{1}{3}$ar3=31​ in equation $(1)$(1) we get,

$$\frac{{r}^{2}}{3}+\frac{{r}^{4}}{3}=2$$3r2​+3r4​=2

$$\Rightarrow {r}^{4}+{r}^{2}=6$$⇒r4+r2=6

$$\Rightarrow ({r}^{2}+3)({r}^{2}-2)=0$$⇒(r2+3)(r2−2)=0

$$\Rightarrow {r}^{2}=2$$⇒r2=2

$$a{r}^{3}=\frac{1}{3}\Rightarrow ar\cdot 2=\frac{1}{3}\Rightarrow ar=\frac{1}{6}$$ar3=31​⇒ar⋅2=31​⇒ar=61​

Now finding the value of $6({a}_{2}+{a}_{4})({a}_{4}+{a}_{6})$6(a2​+a4​)(a4​+a6​), we get

$$=6(ar+a{r}^{3})(a{r}^{3}+a{r}^{5})$$=6(ar+ar3)(ar3+ar5)

$$=6(\frac{1}{6}+\frac{1}{3})(\frac{1}{3}+\frac{2}{3})$$=6(61​+31​)(31​+32​)

$$=6\cdot \frac{1}{2}\cdot 1=3$$=6⋅21​⋅1=3