LetA_1,A_2,A_3,… be an increasing geometric progression of positive real numbers. If A_1 A_3 A_5 A_7= 1 1296 and A_2+A_4= 7 36, then, the value of…

Given

$${A}_{1}{A}_{3}{A}_{5}{A}_{7}=\frac{1}{1296}$$A1​A3​A5​A7​=12961​

Expanding using ${A}_{n}={A}_{1}{r}^{n-1}$An​=A1​rn−1 formula we get,

$$\Rightarrow ({A}_{1})({A}_{1}{r}^{2})({A}_{1}{r}^{4})({A}_{1}{r}^{6})=\frac{1}{1296}$$⇒(A1​)(A1​r2)(A1​r4)(A1​r6)=12961​

$$\Rightarrow {A}_{1}^{4}{r}^{12}=\frac{1}{1296}\Rightarrow {A}_{1}{r}^{3}=\frac{1}{6}...(i)$$⇒A14​r12=12961​⇒A1​r3=61​...(i)

Also given ,

$${A}_{2}+{A}_{4}=\frac{7}{36}\Rightarrow {A}_{1}r+{A}_{1}{r}^{3}=\frac{7}{36}$$A2​+A4​=367​⇒A1​r+A1​r3=367​

$$\Rightarrow {A}_{1}r+\frac{1}{6}=\frac{7}{36}$$⇒A1​r+61​=367​

$$\Rightarrow {A}_{1}r=\frac{1}{36}...(\mathrm{ii})$$⇒A1​r=361​...(ii)

Dividing equation (i) from equation (ii) we get, ${r}^{2}=6$r2=6

Now ${A}_{6}+{A}_{8}+{A}_{10}={A}_{1}{r}^{5}+{A}_{1}{r}^{7}+{A}_{1}{r}^{9}$A6​+A8​+A10​=A1​r5+A1​r7+A1​r9

$={A}_{1}{r}^{3}[{r}^{2}+{r}^{4}+{r}^{6}]$=A1​r3[r2+r4+r6] $=\frac{1}{6}[6+36+216]$=61​[6+36+216] $=43$=43