Question

The product of first three terms of a GP is $1000$ .If $6$ is added to its second term and $7$ added to its third term, the term become in AP. Find the GP.

Answer 1

Step 1: Finding the value of $a$ and $r$

Let the first three terms of the given GP be $\frac{a}{r},a$ and $ar$

According to the question, we get

Product of the first three terms of GP $=1000$
$\Rightarrow \frac{a}{r}\times a\times ar=1000$
$\Rightarrow a^3=(10{)}^3$
$\therefore a=10$
It is given that $\frac{a}{r},a+6,ar+7$ are in AP.
$\therefore 2(a+6)=\frac{a}{r}+ar+7$
$[\because \mathrm{If}\mathrm{x},\mathrm{y},\mathrm{z}$ are in AP, then $2\mathrm{y}=\mathrm{x}+\mathrm{z}]$
$\Rightarrow 2(10+6)=\frac{10}{r}+10r+7[\because a=10]$
$\Rightarrow 32=\frac{10}{r}+10r+7$
$\Rightarrow 25=\frac{10}{r}+10r$

$\Rightarrow 5=\frac{2+2r^2}{r}$
$\Rightarrow 2r^2-5r+2=0$
$\Rightarrow 2r^2-r-4r+2=0$
$\Rightarrow r(2r-1)-2(2r-1)=0$
$\Rightarrow (2r-1)(r-2)=0$
$\Rightarrow r=2,\frac{1}{2}$

Step 2: Finding the terms of GP

Now, we have
$a=10$ and

$r=2$ or

$r=\frac{1}{2}$
Taking $a=10$ and $r=2$ we get
The required terms are
$\left\{\frac{10}{2},10,10\times 2\right\}$
i.e. $\left\{5,10,20\right\}$
And also taking $a=10$ and $r=\frac{1}{2}$ we get
The required terms are
$\left\{\frac{10}{{\displaystyle \frac{1}{2}}},10,10\times \frac{1}{2}\right\}$

i.e., $\left\{20,10,5\right\}$
Hence, the first three terms of GP are $5,10,20$ for $r=2$ and $20,10,5$ for $r=\frac{1}{2}$