Question

the first term of a gp is 243 and the 5th term is 48 find the sum of its first 4 terms

Answer 1

Question :

The first term of a gp is 243 and the 5th term is 48 .Find the sum of its first 4 terms.

Theory :

The sequence $x_{1},x_{2},x_{3}.....x_{n}$ is is called a geometric progression(GP).If

$\sf \dfrac{x _{2} }{x_{1}} = \dfrac{x_{3} }{x_{2} } = \dfrac{x_{n}}{x _{n - 1} }$,where none of $x_{1},x_{2},x_{3}.....x_{n}$ is zero.

⇒In genral $\frac{x_{n+1}}{x_{n}}$= constant(r) .This constant ratio (r) called common ratio .

Formula's used:

1) Genral term of a GP (nth term )

$\bf\:T_{n}=ar{}^{n-1}$

2) Sum of n terms of a GP

The sum of first n terms of an GP is given by

$\sf\:S_{n}=a(\dfrac{1-r{}^{n}}{1-r})$,when r<1

or $\sf\:S_{n}=a(\dfrac{r{}^{n}-1}{r-1})$,when r>1

Solution :

Given : First term ,a= 243

$a_{5}=48$

We have to find sum of its first 4 terms.

________________________

$\sf \: a _{5} = ar {}^{4}$

$\implies48 = 243r {}^{4}$

$\implies \sf r {}^{4} = \frac{ 48}{243}$

$\implies \sf r {}^{4} = \frac{16}{81} = ( \frac{2}{3} ) {}^{4}$

$\implies \sf r = \frac{2}{3}$

Here r<1 , therefore

$\bf\:S_{n}=a(\dfrac{1-r{}^{n}}{1-r})$

⇒$S_{4}=a(\dfrac{1-r{}^{4}}{1-r})$

$= \sf 243 \times ( \dfrac{1 - (\frac{2}{3}) {}^{4} }{1 - \frac{2}{3} } )$

$= \sf 243 \times ( \frac{1 - \frac{16}{81} }{1 - \frac{2}{3} } )$

$= 243( \frac{ \frac{81 - 16}{81} }{ \frac{3 - 2}{3} } )$

$= 243 \times ( \frac{65 \times 3}{81} )$

$= \sf 243 \times ( \frac{65}{27})$

$= \sf9 \times 65$

$= 585$

Hence ,The sum of first 4 terms of a GP is 585 .