SUM OF THE TERMS IN A GEOMETRIC SEQUENCE

Question 1 :

Find the sum of first n terms of the G.P

(i) 5, -3, 9/5, -27/25, ...............

Solution :

a = 5, r = t₂/t₁  =  -3/5  ≠ 1

Sn  =   a(rⁿ - 1)/(r - 1)

   =   5((-3/5)ⁿ - 1)/((-3/5) - 1)

   =   5((-3/5)ⁿ - 1)/((-3-5)/5)

   =   5((-3/5)ⁿ - 1)/(-8/5)

   =   (-25/8)((-3/5)ⁿ - 1)

Sn  =   (25/8)(1  - (-3/5)ⁿ)

(ii) 256, 64, 16,…

Solution :

a = 256, r = t₂/t₁  =  64/256  =  1/4 ≠ 1

Sn  =   a(rⁿ - 1)/(r - 1)

   =   256((1/4)ⁿ - 1)/((1/4) - 1)

   =   256((1/4)ⁿ - 1)/(-3/4)

   =   256(-4/3)((1/4)ⁿ - 1)

S_n  =  (-1024/3)((1/4)ⁿ - 1)

S_n  =  (1024/3)(1 - (1/4)ⁿ)

Question 2 :

Find the sum of first six terms of the G.P. 5, 15, 45, …

Solution :

n = 6, a = 5, r = 15/5  =  3  ≠ 1

Sn  =   a(rⁿ - 1)/(r - 1)

   =   5(3⁶ - 1)/(3 - 1)

=  5(729 - 1)/2

=  5(728/2)

=  1820

Question 3 :

Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.

Solution :

S₆  =  46872

Sn  =   a(rⁿ - 1)/(r - 1)

r = 5 and n = 6

S₆  =   a(5⁶ - 1)/(5 - 1)

46872(4)  =  a(5⁶ - 1)

46872(4) / 15624  =  a

a  =  12

Hence the first term is 12.

Question 4 :

Find the sum to infinity of

(i) 9 + 3 + 1 + .....

Solution :

Sum of infinite series :

S_∞  =  a/(1-r)

a = 9, r = 3/9  =  1/3

S_∞   =  9/(1-(1/3))

=  9/(2/3)

=  27/2

Hence the required sum is 27/2.

(ii) 21 + 14 + 28/3 + .............

Solution :

a = 21, r = 14/21  =  2/3

Sum of infinite series :

S_∞  =  a/(1-r)

S_∞   =  21/(1-(2/3))

=  21/(1/3)

=  63

Hence the required sum is 63.

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.