Question

The sum of three consecutive terms in a geometric progression is $14$.

If $1$ is added to the first and the second terms and $1$ is subtracted from the third, the resulting new terms are in arithmetic progression.

Then the lowest of original term is

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is B

$2$

Explanation of the correct option:

Let $\mathrm{a},\mathrm{ar},{\mathrm{ar}}^2$ be the three consecutive terms in G.P.

$$\begin{gathered} \mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^2=14 \\ \Rightarrow \mathrm{a}(1+\mathrm{r}+{\mathrm{r}}^2)=14...\left(1\right) \end{gathered}$$

Since $\mathrm{a}+1,\mathrm{ar}+1,{\mathrm{ar}}^2-1$ is an A.P.

$\begin{array}{rcl}2\left(\mathrm{ar}+1\right)& =& \mathrm{a}+1+{\mathrm{ar}}^2-1\\ & \Rightarrow & 2\left(\mathrm{ar}+1\right)=\mathrm{a}+{\mathrm{ar}}^2\\ & \Rightarrow & 2\left(\mathrm{ar}+1\right)=14-\mathrm{ar}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}3\mathrm{ar}& =& 12\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\mathrm{ar}& =& 4\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\mathrm{r}& =& \frac{4}{\mathrm{a}}\end{array}$

Substitute the value of $\mathrm{r}$ in equation $\left(1\right)$,

$\mathrm{a}\left(1+\frac{4}{\mathrm{a}}+\frac{16}{{\mathrm{a}}^2}\right)=14$

$\Rightarrow$ $\mathrm{a}+4+\frac{16}{\mathrm{a}}=14$

$\Rightarrow$ ${\mathrm{a}}^2+4\mathrm{a}+16=14\mathrm{a}$

$\Rightarrow$${\mathrm{a}}^2-10\mathrm{a}+16=0$

$\Rightarrow$ $(\mathrm{a}-8)(\mathrm{a}-2)=0$

Thus, $\mathrm{a}=8$ or $\mathrm{a}=2$.

If $\mathrm{a}=8$ then $\mathrm{r}=\frac{1}{2}$,

So, the series is $8,4,2.....$

If $\mathrm{a}=2$ then $\mathrm{r}=2$,

So, the series is $2,4,8.....$

Therefore the lowest term is $2$.

Hence, option (B) is the correct option.