# The sume of the infinite geometric series 3/2+9/16+27/128+81/1024 + ….. is?

How do I do this problem ????

• 3/2+9/16+27/128+81/1024 + .....

Let S be the sum of the given series. Then

S = (3/2) + (9/16) + (27/128) + (81/1024) + .......... inf.

Here the common ratio (r) = Any term divided by the term that follows it.

For example choose the term (81/1024) the following term is (27/128) hence -

Common ratio (r) = (81/1024) divided by (27/128)

=> r = 3/8 This common ratio is less than one, hence the sum to infinite terms can be found out. ( Note that exact sum of a GP having infinite number of terms can be obtained iff, ' r ' is a fraction.)

Hence the required sum : -

S = [ a / ( 1 - r ) ] , where a is first term of the series,

=> S = [ 3/2 divided by ( 1 - 3/8 ) ]

=> S = ( 3/2 ) divided by ( 5/8 )

=> S = 12/5 ..................................... Answer

PKT

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• a = 3/2

r = 3/8

S = a(1 - r^n)/(1 - r)

= (3/2)[1 - (3/8)^infinite]/[1 - 3/8]

(3/8)^infinite = 0 since the value becomes smaller and smaller as the power is increased.

So

S = a / (1 - r) = 3/2 * 8/5 = 24/10

Hope this helps.

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