sum of infinite terms of G.P is 12 and sum of their square is 45 find series
paridhigupta1234:
here its not 12 its 15 sorry typing mistake hehe
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hello pari :-)
as we know that sum of infinite terms of G.P is = 12 which means a/(1-r) = > 15
a² + a²r² + a²r^4 .............................infinite = > 45
which means
a²/ (1-r²) = > 45
a ×a / (1-r) (1+r) = > 45
and here a/ (1-r) = > 15 putting values we get
a15/ (1+r) = > 45
we get a/(r+1) = > 3
we get a = > 3(r+1)
and we know that a/ (1-r) = > 15 putting a = > 3( r+1) we get
3(r+1)/(1-r) = > 15
we get r+1 = > 5( 1 - r)
we get r = > 2/3
and we know that a = > 3 ( r+1)
a= > 3( 2/3 + 1 )
a = > 5 which means
a + ar + ar ^2 ................................................
putting r and a we get
5 + 10/3 + 20/9 ................................. is our required answer
hopes this will help you
@ regards gopal khandelwal doing b-tech from IIT ROORKEY
as we know that sum of infinite terms of G.P is = 12 which means a/(1-r) = > 15
a² + a²r² + a²r^4 .............................infinite = > 45
which means
a²/ (1-r²) = > 45
a ×a / (1-r) (1+r) = > 45
and here a/ (1-r) = > 15 putting values we get
a15/ (1+r) = > 45
we get a/(r+1) = > 3
we get a = > 3(r+1)
and we know that a/ (1-r) = > 15 putting a = > 3( r+1) we get
3(r+1)/(1-r) = > 15
we get r+1 = > 5( 1 - r)
we get r = > 2/3
and we know that a = > 3 ( r+1)
a= > 3( 2/3 + 1 )
a = > 5 which means
a + ar + ar ^2 ................................................
putting r and a we get
5 + 10/3 + 20/9 ................................. is our required answer
hopes this will help you
@ regards gopal khandelwal doing b-tech from IIT ROORKEY
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