FIND THE SUM OF N TERMS FROM THE GIVEN TERMS OF THE GEOMETRIC SERIES

Example 1 :

The second term of the geometric series is 3 and the common ratio is 4/5. Find the sum of first 23 consecutive terms in the given geometric series.

Solution :

Second term (t₂)  =  3

ar  =  3    r  =  4/5 < 1   and n  =  23

a(4/5)  =  3

a  =  (3 ⋅ 5)/4

a  =  15/4 

Sum of first 23 terms = a (1 - rⁿ)/(1 - r)

  =  15/4 (1 - (4/5)²³)/(1 - (4/5))

  =  15/4 [ (1 - (4/5)²³)/(1/5) ]

  =  (15/4) ⋅ (5/1) (1-(4/5)^23)

  =  (75/4) (1-(4/5)^23)

Example 2 :

Suppose that five people are ill during the first week of an epidemic and each sick person spreads the contagious disease to four other people by the end of the second week and so on. By the end of 15^th week, how many people will be affected by the epidemic?

Solution :

By writing the given data as sequence, we get

5, 5(4), 5(4)², ..............

5, 20, 80, ..............

Since we find total number of people affected by the epidemic the above sequence is going to be changed as 

5 + 20 + 80 + ..............

here a  =  5  r  =  20/5 ==> 4 and n = 15

S_n  =  a (rⁿ-1)/(r-1)

S₁₅  =  5 (4¹⁵ - 1)/(4 - 1)

S₁₅  =  (5/3) (4¹⁵ - 1)

Example 3 :

A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have 1000 mangoes at once or he could get 1 mango on the first day, 2 on the second day, 4 on the third day, 8 mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes ?

Solution :

By choosing the first option, he will get 1000 mangoes. 

To find the total number of mangoes that he is going to get by choosing the second option, we should find the sum of first 10 terms of the series.

Writing the second option as series, we get

1 + 2 + 4 + 8 + .......  10 days

Here a = 1       r = 2 > 1

s_n = a (rⁿ - 1) / (r - 1)

  =  1 (2¹⁰ - 1)/(2 - 1)  

  =  1024 - 1

  =  1023

By choosing the second way, he will get 1023 mangoes.

Therefore the boy should choose the second way to get the maximum number of mangoes.

Example 4 :

A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 9 and the sum of the last two terms is 36. Find the series.

Solution :

Let a, ar, ar²  and ar³ be the first four terms of geometric series

Sum of the first two terms  =  9

a + ar  =  9

a(1 + r)  =  9  ----(1)

Sum of the last two terms = 36

ar² + ar³  =  36

ar² (1 + r)  =  36    --- (2)

Substitute a (1 + r) = 9 in (2), we get

r² (9)  =  36

r²  =  36/9

r²  =  4

r  =  √4

r  =  ± 2

r  =  -2 is not admissible. So, the value of r is 2.

By applying the value of r in (1), we get

a (1 + 2)  =  9

a(3)  =  9

a  =  3

By applying the values of a and r, we get

3 + 3(2) + 3(2)² + 3(2)³+ .........

Therefore, the series  is 

3 + 6 + 12 + 24 + ......

Example 5 :

A person borrows $8000 at 2.76% simple interest per annum. The principal and the interest are to be paid in the 10 monthly instalments. If each instalment is double the preceding one, find the value of the first and the last instalment.

Solution :

Interest to be paid = Simple interest = PNR/100

P = 8000, r = 2.76

Number of months of interest to be paid = 10

Converting into number of years = 10/12

= (8000 x 2.76 x 10) / (100 x 12)

= 184

Total amount to be paid with interest = 8000 + 184

= 8184

This amount should be paid into instalments, but it is not equal instalments. Each instalment is double the preceding one. By creating the instalments as sequence, it must be a geometric sequence.

s_n = 8184

a (rⁿ - 1) / (r - 1) = 8184

n = 10, r = 2

a (2¹⁰ - 1) / (2 - 1) = 8184

a (1024 - 1)/1  = 8184

a (1023)  = 8184

a = 8184/1023

a = 8

Amount to be paid for the last instalment = 10^th instalment.

t₁₀ = ar⁹

t₁₀ = 8(2)⁹

= 8(512)

t₁₀ = 4096

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