Question

the sum of 3 terms of gp is 7 and that of 6 terms is 63. find gp​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that,

↝ First term of GP series, is a

and

↝ Common ratio of GP series, is r.

Now,

↝ Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a( {r}^{n} - 1) }{r - 1} \: \: provided \: that \: r \: \ne \: 1 }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of GP.

a is the first term of the sequence.

n is the no. of terms.

r is the common ratio.

Now,

Given that,

↝ Sum of first three terms of GP = 7

$\red{\rm :\longmapsto\:S_3 = 7}$

$\bf :\longmapsto\:\dfrac{a( {r}^{3} - 1) }{r - 1} = 7 - - - (1)$

Also, given that,

↝ Sum of first 6 terms = 63

$\red{\rm :\longmapsto\:S_6 = 63}$

$\bf :\longmapsto\:\dfrac{a( {r}^{6} - 1) }{r - 1} = 63 - - - (2)$

Now, Divide equation (2) by (1), we get

$\bf :\longmapsto\:\dfrac{( {r}^{6} - 1) }{ {r}^{3} - 1} = 9$

can be rewritten as

$\bf :\longmapsto\:\dfrac{(( {r}^{3}) {}^{2} - 1) }{ {r}^{3} - 1} = 9$

$\bf :\longmapsto\:\dfrac{( {r}^{3} - 1)( {r}^{3} + 1)}{ {r}^{3} - 1} = 9$

$\rm :\longmapsto\: {r}^{3} + 1 = 9$

$\rm :\longmapsto\: {r}^{3} = 9 - 1$

$\rm :\longmapsto\: {r}^{3} = 8$

$\rm :\longmapsto\: {r}^{3} = {2}^{3}$

$\bf\implies \:r = 2$

On Substituting r = 2, in equation (1) we get

$\bf :\longmapsto\:\dfrac{a( {2}^{3} - 1) }{2 - 1} = 7$

$\bf :\longmapsto\:\dfrac{a( {2}^{3} - 1) }{2 - 1} = 7$

$\bf :\longmapsto\:\dfrac{a( 8 - 1) }{1} = 7$

$\bf\implies \:a = 1$

So, Required GP series is

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{\boxed{ \bf{ \: 1, \: 2, \: 4, \: 8, \: 16, \: - - - - }}}$

Answer 2

Step-by-step explanation:

Let the first term og gp series is a

and common ratio of gp series, is r

Now, we know that :-

$\bold{ \: S_n = \frac{a( {r}^{n} - 1)}{r - 1} }$

According to the question :-

The Sum of 3 terms of gp = 7

$\rightarrow \: S_3 = 7$

$\therefore \: \: \bold{ \frac{a( {r}^{3} - 1)}{r - 1} } = 7 - - - (i)$

and sum of first 6 terms of gp = 63

$\rightarrow \: S_6 = 63$

$\therefore \: \: \bold{ \frac{a( {r}^{6} - 1)}{r - 1} } = 63 - - - (ii)$

Now, Divide equation (ii) by (i), we get

$\frac{a( {r}^{n} - 1) }{r - 1} \div \frac{a( {r}^{n} - 1) }{r - 1} = 63 \div 7$

$\frac{ \cancel{a}( {r}^{6} - 1) }{ \cancel{r - 1}} \times \frac{ \cancel{r - 1} }{ \cancel{a}( {r}^{3} - 1)} = \cancel{63} \times \frac{1}{ \cancel7}$

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

$\frac{ (({r}^{3})^{2} - 1) }{ ({r}^{3} - 1)} = 9$

$\frac{( {r}^{3} + 1) \cancel{( {r}^{3} - 1)}}{ \cancel{( {r}^{3} - 1)}} = 9$

${r}^{3} + 1 = 9$

${r}^{3} = 9 - 1$

${r}^{3} = 8$

${r}^{3} = {(2})^{3}$

$r = 2$

On Substituting r = 2, in equation (i), we get

${ \frac{a( {2}^{3} - 1)}{2- 1} } = 7$

${ \frac{a( 8 - 1)}{1} } = 7$

${ 7a} = 7$

$a = \frac{ \cancel7}{ \cancel7}$

$a = 1$