# WORD PROBLEMS FINITE GEOMETRIC SERIES

Problem 1 :

A man repays an amount of Rs.3250 by paying Rs.20 in the first month and then increases the payment by Rs.15 per month. How long will it take him to clear the amount?

Solution :

Sn  =  3250

a  =  20, d  =  15

(n/2)[2a + (n-1)d]  =  3250

(n/2)[2(20) + (n-1)(15)]  =  3250

(n/2)[40+15n-15]  =  3250

(n/2)[40+15n-15]  =  3250

(n/2)[25 + 15n]  =  3250

n(5+3n)  =  3250(2)/5

5n + 3n2  =  1300

3n2 + 5n  - 1300 = 0

(n - 20) (3n + 65)  =  0

n - 20  =  0               3n + 65  =  0

n  =  20                       n  =  -65/3

So, the required number of moths to clear the amount is 20.

Problem 2 :

In a race, 20 balls are placed in a line at intervals of 4 meters, with the first ball 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?

Solution :

Distances traveled to pic up the balls are 48,56,64,...

a = 48, d = 8

S20  =  (20/2) [248 + (20 - 1) ⋅ 8]

=  10 [96 + 152]

=  2480 metre

Problem 3 :

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

Solution :

Given that there were 30 bacteria present originally.

a  =  30

Also given that the number doubles every one hour.

After one hour the count  =  2(30)  =  60

Since the count doubles every hour it forms G.P with r=2

a, ar, ar2, ...................

The count at the end of the 2nd hour

30 (2)2  =  120

The count at the end of the 4th hour

30 (2)4  =  480

The count at the end of the nth hour

30 (2)n

Problem 4 :

What will Rs.500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Solution :

We use the formula

A  =  P[1 + r/100)]n

Here P  =  Principal

r  =  rate of interest  =  10%

t  =  time  =  10 years

A  =  Amount

Amount  =  500 [1 + (10/100)]10

=  500 [1 + (1/10)]10

=  500(11/10)10

=  500(1.1)10

Problem 5 :

In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over 150,000 units.

Solution :

a  =  5, r  =  2 and tn > 150000. To find n

tn  =  arn-1

arn-1  =  150000

5(2)n-1  =  150000

2n-1  =  150000/5

2n-1  =  30000

215  =  32768 > 30000

214  =  16384

So, on the 15th day it will grow over 150,000 units.

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