Question

use the formula Sn=a(r^n-1)/r-1 to find a1 and Sn if an=1000,r=10,n=7

Answer 1

ANSWER :–

$\\ \longrightarrow \: \: { \red { \boxed{ \bold{ a_{1} = {10 }^{ - 3} }}}}$

$\\ \longrightarrow\: \: { \red{ \boxed{ \bold{ s_{n} = 1111.111 }}}}$

EXPLANATION :–

GIVEN :–

$\\ \longrightarrow \: \: { \bold{ n \th \: \: term(a_{n}) = 1000 }}$

$\\ \longrightarrow \: \: { \bold{ common \: \: ratio(r) = 10 }}$

$\\ \longrightarrow \: \: { \bold{ number \: \: of \: \: terms (n) = 7 }}$

TO FIND :–

$\\ \longrightarrow \: \: { \bold{ a_{1} (first \: \: term) \: \: , \: \: S_{n}(sum \: \: of \: \: n \: \: terms) }}$

SOLUTION :–

▪︎ We know that nth term G.P. is –

$\\ \implies \: \: { \boxed{ \red{ \bold{ a_{n} = (a_{1}) \: {r}^{n - 1} }}}}$

• Now put the values –

$\\ \implies \: \: { \bold{ 1000 = (a_{1}) \: {10 }^{7 - 1} }}$

$\\ \implies \: \: { \bold{ 1000 = (a_{1}) \: {10 }^{6} }}$

$\\ \implies \: \: { \boxed{ \bold{ a_{1} = {10 }^{ - 3} }}}$

▪︎ We also know that –

$\\ \implies \: \: { \boxed{ \red{ \bold{ s_{n} = \frac{a ({r}^{n } - 1)}{r - 1} }}}}$

• Let's put the values –

$\\ \implies \: \: { \bold{ s_{n} = \frac{( {10}^{ - 3})({10}^{7 } - 1)}{10 - 1} }}$

$\\ \implies \: \: { \bold{ s_{n} = \frac{( {10}^{ - 3})(10000000 - 1)}{9} }}$

$\\ \implies \: \: { \bold{ s_{n} = \frac{( {10}^{ - 3})( \cancel {9999999})}{ \cancel9} }}$

$\\ \implies \: \: { \bold{ s_{n} = ({10}^{ - 3} )(1111111) }}$

$\\ \implies \: \: { \boxed{ \bold{ s_{n} = 1,111.111 }}}$

Answer 2

Given ,

• Last term of GP (an) = (10)³
• Common ratio (r) = 10
• No. of terms (n) = 7

We know that , the first n terms of an GP is given by

$\star \: \: \sf a_{n} = a {r}^{(n - 1)}$

Thus ,

$\sf \Rightarrow {(10)}^{3} = a {(10)}^{7 - 1} \\ \\ \sf \Rightarrow {(10)}^{3} = a {(10)}^{6} \\ \\ \sf \Rightarrow a = \frac{1}{ {(10)}^{3} } \\ \\ \sf \Rightarrow a = {(10)}^{ - 3}$

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

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

Thus ,

$\sf \Rightarrow S_{7} = \frac{ {(10)}^{ - 3} ( {10}^{7} - 1)}{10- 1} \\ \\\sf \Rightarrow S_{7} = \frac{{(10)}^{ - 3} \times \cancel{9999999}}{ \cancel{9}} \\ \\\sf \Rightarrow S_{7} = 1111111 \times {(10)}^{ - 3} \\ \\\sf \Rightarrow S_{7} = 1111.111</p><p></p><p>$

$\therefore \sf \underline{ \bold{a = {(10)}^{ - 3} \: \: and \: \: S_{7} = 1111.111}}$