Question

Let $S_n\left(x\right)={\log }_{a^{\frac{1}{2}}}x+{\log }_{a^{\frac{1}{3}}}x+{\log }_{a^{\frac{1}{6}}}x+{\log }_{a^{\frac{1}{11}}}x........$ up to $n-$terms, where $a>1$ If $S_{24}\left(x\right)=1093and{S}_{12}\left(2x\right)=265$ then value of $a$ is equal to

Answer 1

Determine the value of $a$

On solving the given function, we get:

$$\begin{gathered} S_n\left(x\right)={\log }_{a^{\frac{1}{2}}}x+{\log }_{a^{\frac{1}{3}}}x+{\log }_{a^{\frac{1}{6}}}x+{\log }_{a^{\frac{1}{11}}}x........ \\ {S}_n\left(x\right)={\log }_a{x}^2+{\log }_a{x}^3+{\log }_a{x}^6+{\log }_a{x}^{11}........ \\ {S}_n\left(x\right)={\log }_a{x}^{2+3+6+11}\left[\because \mathrm{loga}+\mathrm{logb}=\mathrm{logab}\right] \end{gathered}$$

Now,$T_n=2+{(n-1)}^2$

$$\begin{gathered} S_n=\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}{T}_n \\ =\sum _{\mathrm{n}=1}2+{(\mathrm{n}-1)}^2=2\mathrm{n}+\frac{(\mathrm{n}-1)\left(\mathrm{n}\right)(2\mathrm{n}-1)}{6} \end{gathered}$$

Therefore, we can write

$$\begin{gathered} S_n\left(x\right)={\log }_a{x}^{\left[2\mathrm{n}+\frac{(\mathrm{n}-1)\left(\mathrm{n}\right)(2\mathrm{n}-1)}{6}\right]} \\ \text{Now}, \\ {S}_{24}\left(x\right)={\log }_a{x}^{{}^{\left[48+\frac{23.24.47}{6}\right]}} \\ \Rightarrow 1093=4372{\log }_{a}x \\ \Rightarrow {\log }_{a}x=\frac{1}{4}.................\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} S_{12}\left(2x\right)=265 \\ {\log }_{a}2x^{\left[24+\frac{11.12.23}{6}\right]}=265 \\ \Rightarrow 530{\log }_{a}2x=265 \\ \Rightarrow {\log }_{a}2x=\frac{1}{2}.............\left(2\right) \end{gathered}$$

On subtracting $\left(2\right)-\left(1\right)$, we get:

$$\begin{gathered} {\log }_{a}2x-{\log }_{a}x=\frac{1}{2}-\frac{1}{4} \\ {\log }_{a}2=\frac{1}{4} \\ a=16 \end{gathered}$$

Therefore the value of $ais16.$