Question

If third and fourth terms of an infinite geometric sequence are `27` and `20 1/4`, find its tenth term and sum up to infinity?

Answer 1

Tenth term is `59049/16384` sum of infinite series is `192`

Explanation:

Let the first term of geometric sequence be `a` and common ratio be `r`. Then third term is `ar^2=27` and fourth term is `ar^3=20 1/4=81/4`. Dividing latter by former, we get

`(ar^3)/(ar^2)=(81/4)/27` or `r=81/4xx1/27=3/4`

and hence `axx(3/4)^2=27` or `a=27xx4/3xx4/3=48`

and tenth term is `48xx(3/4)^9=3xx4^2xx3^9/4^9=3^10/4^7`

= `59049/16384`

As `|r|<1`, sum of infinite series is `a/(1-r)=48/(1/4)=192`

Answer 2

Tenth term `a_10 = color (purple)(3.6041)`

`S_oo = a / (1- r) = color(green)(192)`

Explanation:

Nth term of a G. S `a_n = a * r^(n-1)`

Third term `a_3 = a * r^2 = 27` Eqn (1)

Fourth term `a_4 = a * r^(3) = 20(1/4)` Eqn (2)

Dividing Eqn (2) by (1),

`(a r^3) / (a r^2) = r = (20(1/4)) / 27 =( 3/4)`

Substituting the value of ‘r’ in Eqn (1),

`a r^2 = 27 = a * (3/4)^2`

`a = 27 / (3/4)^2 = 48`

Tenth term `a_10 = a * r^9 = 48 * (3/4)^9 = color(purple)(3.6041)`

Sum of geometric sequence of infinite series with ‘r’ less than 1 is given by the formula,

`S_oo = a / (1- r) = 48 / (1 - 3/4) = color(green)(192)`