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

2 Answers

• 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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