Question

In a Geometric progression first term is 2, common ratio is 3 and Sn= 6560. Find the value of n.​

Answer 1

Answer:

$n \: = \: 8$

Step-by-step explanation:

Given---->

$for \: a \: geoetric \: progression \: first \: term \: is \: 2 \\ \: common \: ratio \: is \: 3 \: and \: sum \: of \: n \: terms \: is \: 6560$

To find---->

$value \: of \: n \:$

Concept used---->

$sum \: of \: n \: terms \: = \dfrac{a( {r}^{n} - 1)}{(r - 1)}$

Solution---->

$a = 2 \: \\ r \: = 3 \\ sum \: of \: n \: terms \: = 6560$

$now \:$

$sum \: of \: n \: terms \: = \dfrac{a( {r}^{n} - 1) }{(r - 1)}$

$= > 6560 \: = \dfrac{2 \: ( \: {3}^{n} - 1 \: ) }{( \: 3 - 1 \: )}$

$= > 6560 \: = \dfrac{2 \: ( \: {3}^{n} \: - \: 1 \: ) }{2}$

$= > 6560 \: = {3}^{n} \: - \: 1$

$= > {3}^{n} \: = 6560 \: + \: 1$

$= > {3}^{n} \: = 6561$

$= > {3}^{n} \: = \: {3}^{8}$

$comparing \: exonent \: from \: both \: sides \:$

$= > n \: = \: 8$

Additional information---->

1)

$nth \: term \: of \: a \: gp \: \\ = a \: {r}^{n - 1}$

2)

$sum \: of \: infinite \: terms \: of \: gp \\ = \dfrac{a}{( \: 1 - r \: )}$