Question

2. In a geometric series, e^(m+1),(m + 1)e^(m+1 )and (4m + 1)e^(m+1) are consecutive terms.
a) Calculate the value of m and the common ratio.
b) Hence, if the third term of the series is em+1, find the sum of the first 6 terms of the series.​

Answer 1

Given:

$e^{(m+1)},(m+1)e^{(m+1)}$ and $(4m+1)e^{(m+1)}$ are consecutive terms in a GP.

Third term of the series = $e^{(m+1)}$

To find:

Value of $m$ and common ratio.

Sum of first 6 terms.

Solution:

Since it is a geometric series, let the series be in form $a,ar,ar^{2},ar^{3}...ar^{n}$

where $a$ is the first term and $r$ is the common ratio.

The ratio between any two consecutive terms is common ratio, i.e., the common ratio between first two consecutive terms is equal to the common ratio of the next two consecutive terms.

Hence,

$\frac{(m+1)e^{m+1}}{e^{m+1}}=\frac{(4m+1)e^{m+1}}{(m+1)e^{m+1}}$

$e^{m+1}$ will be cancelled out from both sides.

$m+1=\frac{4m+1}{m+1}$

$(m+1)^{2}=4m+1$

$m^{2} +2m+1=4m+1$

$m^{2} +2m+1-4m-1=0$

$m^{2}-2m=0$

$m(m-2)=0$

$m=0$ or $m-2=0$

$m=0$ or $m=2$

If $m=0$, the consecutive terms will be:

$e^{m+1}=e^{0+1}=e$

$(m+1)e^{m+1}=(0+1)e^{0+1}=e$

$(4m+1)e^{m+1}=(4*0+1)e^{0+1}=e$

Hence, the consecutive terms are $e,e,e$ and it is an infinite series.

If $m=2$, the consecutive terms will be:

$e^{m+1}=e^{2+1}=e^{3}$

$(m+1)e^{m+1}=(2+1)e^{2+1}=3e^{3}$

$(4m+1)e^{m+1}=(4*2+1)e^{2+1}=9e^{3}$

Hence, the consecutive terms are $e^{3},3e^{3},9e^{3}$.

To find the common ratio, divide any two consecutive terms.

$r_{1}=\frac{e}{e}=1$

$r_{2}=\frac{3e^{3}}{e^{3}}=3$

Hence, value of common ratio, $r_{1}=1$ when $m=0$ and $r_{2}=3$ when $m=2$.

b) When $m=2$,

Third term of series = $e^{m+1}=e^{2+1}=e^{3}$

To find the second term, divide third term by common ratio.

Second term = $\frac{e^{3}}{3}$

To find first term, divide second term by common ratio.

First term = $\frac{e^{3}}{9}$

To find fourth term, multiply third term with common ratio.

Fourth term = $3e^{3}$

To find fifth term, multiply fourth term with common ratio.

Fifth term = $9e^{3}$

To find sixth term, multiply fifth term with common ratio.

Sixth term = $27e^{3}$

Hence, the series is $\frac{e^{3}}{9},\frac{e^{3}}{3},e^{3},3e^{3},9e^{3},27e^{3}...$

When $m=0$,

third term = $e^{m+1}=e^{0+1}=e$

Second term = $e$

First term = $e$

All the terms will be $e$. Hence, the series is an infinite series given by:

$e,e,e,e,e...$

The sum of $n$ terms in a GP is given by:

$S_{n}=\frac{a(1-r^{n})}{1-r}$ when $r\neq 1$

$S_{n}=an$ when $r=1$

Here, we need to find the sum of first 6 terms. So, $n=6$

For the series, $\frac{e^{3}}{9},\frac{e^{3}}{3},e^{3},3e^{3},9e^{3},27e^{3}...$, $r=3$ which is greater than 1.

$S_{n}=\frac{a(1-r^{n})}{1-r}$

$S_{6}=\frac{\frac{e^{3}}{9} (1-3^{6})}{1-3}$

$S_{6}=\frac{\frac{e^{3}}{9}*-728 }{-2}=\frac{364}{9}e^{3}$

For the series, $e,e,e,e,e...$, $r=1$

$S_{n}=an$

$S_{6}=a*6=6e$

a) The values of m are $0$ and $2$. When $m=0, r=1$ and when $m=2,r=3$.

b) The sum of first 6 terms of the series when $r=1$ is $S_{6}=\frac{364}{9}e^{3}$ and when $r=3$ is $S_{6}=6e$.