use the formula Sn=a(r^n-1)/r-1 to find a1 and Sn if an=1000,r=10,n=7
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r-1 to find a1 and Sn if an=1000,r=10,n=7
use the formula Sn=a(r^n-1)/r-1 to find a1 and Sn if an=1000,r=10,n=7
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ANSWER :–
[tex] \\ \longrightarrow \: \: { \red { \boxed{ \bold{ a_{1} = {10 }^{ - 3} }}}} \\ [/tex]
[tex] \\ \longrightarrow\: \: { \red{ \boxed{ \bold{ s_{n} = 1111.111 }}}} \\ [/tex]
EXPLANATION :–
GIVEN :–
[tex] \\ \longrightarrow \: \: { \bold{ n \th \: \: term(a_{n}) = 1000 }} \\ [/tex]
[tex] \\ \longrightarrow \: \: { \bold{ common \: \: ratio(r) = 10 }} \\ [/tex]
[tex] \\ \longrightarrow \: \: { \bold{ number \: \: of \: \: terms (n) = 7 }} \\ [/tex]
TO FIND :–
[tex] \\ \longrightarrow \: \: { \bold{ a_{1} (first \: \: term) \: \: , \: \: S_{n}(sum \: \: of \: \: n \: \: terms) }} \\ [/tex]
SOLUTION :–
▪︎ We know that nth term G.P. is –
[tex] \\ \implies \: \: { \boxed{ \red{ \bold{ a_{n} = (a_{1}) \: {r}^{n - 1} }}}} \\ [/tex]
• Now put the values –
[tex] \\ \implies \: \: { \bold{ 1000 = (a_{1}) \: {10 }^{7 - 1} }} \\ [/tex]
[tex] \\ \implies \: \: { \bold{ 1000 = (a_{1}) \: {10 }^{6} }} \\ [/tex]
[tex] \\ \implies \: \: { \boxed{ \bold{ a_{1} = {10 }^{ - 3} }}} \\ [/tex]
▪︎ We also know that –
[tex] \\ \implies \: \: { \boxed{ \red{ \bold{ s_{n} = \frac{a ({r}^{n } - 1)}{r - 1} }}}} \\ [/tex]
• Let's put the values –
[tex] \\ \implies \: \: { \bold{ s_{n} = \frac{( {10}^{ - 3})({10}^{7 } - 1)}{10 - 1} }} \\ [/tex]
[tex] \\ \implies \: \: { \bold{ s_{n} = \frac{( {10}^{ - 3})(10000000 - 1)}{9} }} \\ [/tex]
[tex] \\ \implies \: \: { \bold{ s_{n} = \frac{( {10}^{ - 3})( \cancel {9999999})}{ \cancel9} }} \\ [/tex]
[tex] \\ \implies \: \: { \bold{ s_{n} = ({10}^{ - 3} )(1111111) }} \\ [/tex]
[tex] \\ \implies \: \: { \boxed{ \bold{ s_{n} = 1,111.111 }}} \\ [/tex]
Given ,
We know that , the first n terms of an GP is given by
[tex] \star \: \: \sf a_{n} = a {r}^{(n - 1)} [/tex]
Thus ,
[tex] \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} [/tex]
Now , The sum of first n terms of an GP is given by
[tex] \star \: \: \sf S_{n} = \frac{a( {r}^{n} - 1)}{r - 1} [/tex]
Thus ,
[tex]\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
[/tex]
[tex] \therefore \sf \underline{ \bold{a = {(10)}^{ - 3} \: \: and \: \: S_{7} = 1111.111}}[/tex]