Question

If the sequence $\left\{a_n\right\}$ is in GP, such that $\frac{a_4}{a_6}=\frac{1}{4}$ and $a_2+a_5=216$, then $a_1$ is equal to

• A. $12$ or $\frac{108}{7}$
• B. $10$
• C. $7$ or $\frac{54}{7}$
• D. None of these

Answer 1

The correct option is A

$12$ or $\frac{108}{7}$

Step 1: Solve for common ratio

Given that the sequence $\left\{a_n\right\}$ is in GP

We know that $a_n=a.r^{n-1}$ where, $a$ is the first term and $r$ is the common ratio

Given that $\frac{a_4}{a_6}=\frac{1}{4}$

$\begin{array}{crcl}\Rightarrow & \frac{a.r^{4-1}}{a.r^{6-1}}& =& \frac{1}{4}\\ \Rightarrow & \frac{r^3}{r^5}& =& \frac{1}{4}\\ \Rightarrow & \frac{1}{r^2}& =& \frac{1}{4}\\ \Rightarrow & r^2& =& 4\\ \Rightarrow & r& =& \pm 2\end{array}$

Step 2: Solve for the first term

Given that $a_2+a_5=216$

Case 1: $r=2$

$\begin{array}{crcl}\therefore & a{\left(2\right)}^{2-1}+a{\left(2\right)}^{5-1}& =& 216\\ \Rightarrow & 2a+16a& =& 216\\ \Rightarrow & 18a& =& 216\\ \Rightarrow & a& =& \frac{216}{18}\\ \Rightarrow & a& =& 12\end{array}$

Case 2: $r=-2$

$\begin{array}{crcl}\therefore & a{\left(-2\right)}^{2-1}+a{\left(-2\right)}^{5-1}& =& 216\\ \Rightarrow & -2a+16a& =& 216\\ \Rightarrow & 14a& =& 216\\ \Rightarrow & a& =& \frac{216}{14}\\ \Rightarrow & a& =& \frac{108}{7}\end{array}$

Step 3: Solve for the required value

We know that $a_1=a.r^{1-1}$

$$\begin{gathered} =a.r^0 \\ \Rightarrow a_1=a \end{gathered}$$

Therefore $a_1=12$ or $\frac{108}{7}$

Hence, the correct option is option (A) i.e. $12$ or $\frac{108}{7}$.