PROBLEMS ON SPECIAL TYPE OF GEOMETRIC SERIES

Problem 1 :

If the first term of an infinite G.P. is 8 and its sum to infinity is 32/3 then find the common ratio.

Solution :

Since the given sequence is infinite series, the sum of the series  =  a/(1 - r) 

a/(1-r)  =  32/3

a = 8

8/(1-r)  =  32/3

8(3)  =  32(1-r)

24 / 32  =  1 - r

3/4  =  1 - r

r  =  1 - (3/4)  ==>  1/4

Problem 2 :

Find the sum to n terms of the series.

(i) 0.4 + 0.44 + 0.444 +........ to n terms

Solution :

 0.4 + 0.44 + 0.444 +........ to n terms

  =  4(0.1 + 0.11 + 0.111 + ...... to n terms)

  =  4 ⋅ (9/9) (0.1 + 0.11 + 0.111 + ...... to n terms)

  =  (4/9)[0.9 + 0.99 + 0.999 +............n terms]

  =  (4/9)[(1 - 0.1) + (1 - 0.1²) + (1 - 0.1³)  + .......n terms]

  =  (4/9)[(1+1+1........n terms) - [0.1+0.1²+0.1³ + .......n terms]

  =  (4/9)[n - 0.1 [1 - (0.1)ⁿ/(1-0.1)]

  =  (4/9)[n - 0.1 [1 - (0.1)ⁿ/0.9]

  =  (4/9)[n - [(1 - (0.1)ⁿ)/0.9]

(ii) 3 + 33 + 333 + ........... to n terms

Solution :

3 + 33 + 333 + ........... to n terms

  =  3 [1 + 11 + 111 + ...........+ n terms]

  =  3 ⋅ (9/9) (1 + 11 + 111 + ...... to n terms)

  =  (3/9)[9 + 99 + 999 +............n terms]

  =  (1/3)[(10-1) + (100 - 1) + (1000 - 1) +............n terms]

  =(1/3)[(10+100+1000 + .......n terms)-(1+1+1+.........n terms)]

a = 10, r = 100/10  =  10, a = 1 and r = 1

  =  (1/3)[10(10ⁿ - 1)/(10-1))-n]

  =  (1/3)[(10/9)(10ⁿ - 1) - n]

  =  [(10/27)(10ⁿ - 1) - (n/3)]

Problem 3 :

Find the sum of the Geometric series :

3 + 6 + 12 +............+ 1536

Solution :

t_n  =  1536

a  =  3, r = 6/3  =  2

ar^n-1  =  1536

3(2)^n-1  =  1536

2^n-1  =  1536/3

2^n-1  =  512

2^n-1  =  2⁹

n - 1  =  9

n = 10

Now, we have to find the sum of 10 terms.

Sn  =  a(rⁿ - 1)/(r - 1)

S₁₀  =  3(2¹⁰ - 1)/(2 - 1)

  =  3(1024 - 1)

  =  3(1023)

S₁₀  =  3069

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.