# Question #a32a3

Oct 27, 2017

Let the given infinite Geometric series has first term $a$ common ratio $r < 1$

So sum of the infinite series $\frac{a}{1 - r} = 243. \ldots . . \left[1\right]$

And its sum up to 5th term ${S}_{5} = \frac{a \left(1 - {r}^{5}\right)}{1 - r} = 275. \ldots . . \left[2\right]$

Dividing [2] by [1] we get

$1 - {r}^{5} = \frac{275}{243}$

$\implies {r}^{5} = 1 - \frac{275}{243} = - \frac{32}{243} = {\left(- \frac{2}{3}\right)}^{5}$

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

Inserting the value of $r = - \frac{2}{3}$ in [1] we get

$\frac{a}{1 + \frac{2}{3}} = 243$

$\implies a = 243 \times \frac{5}{3} = 405$

Hence

1st term ${t}_{1} = a = 405$

2nd term ${t}_{2} = a \times r = 405 \times \left(- \frac{2}{3}\right) = - 270$

3rd term ${t}_{3} = a \times {r}^{2} = 405 \times {\left(- \frac{2}{3}\right)}^{2} = 180$

4th term ${t}_{4} = a {r}^{3} = 405 \times {\left(- \frac{2}{3}\right)}^{3} = - 120$