Q. 84.0( 9 Votes )

# Rahul’s father wants to deposit some amount of money every year on the day of Rahul’s birthday. On his 1^{st} birth day Rs.100, on his 2^{nd} birth day Rs.300, on his 3rd birth day Rs.600, on his 4th birthday Rs.1000 and so on. What is the amount deposited by his father on Rahul’s 15th birthday.

Answer :

Money deposited by Rahul's father is as follows:

100, 300, 600, 1000

They are in following series

100 + 1 × 0, 100 + 2 × 100, 300 + 3 × 100, 600 + 4 × 100

Multiplication numbers 1, 2, 3 & 4 represent the age Rahul is turning.

The multiplication factor is previous year deposit.

If we say, n is number of years, Rahul is turning and T

T

T

T3 = T

T4 = T3 + n × 100 = T

....

T15 = T

= 1 × 100 + 2 × 100 + 3 × 100 + ........+ 15 × 100

All the terms after T

a = 100 and d = 100 n = 15

Hence T

= n(2a + (n-1)d)/2

= 15(200 + 14 × 100)/2

= 15 × 800

= 12000

100, 300, 600, 1000

They are in following series

100 + 1 × 0, 100 + 2 × 100, 300 + 3 × 100, 600 + 4 × 100

Multiplication numbers 1, 2, 3 & 4 represent the age Rahul is turning.

The multiplication factor is previous year deposit.

If we say, n is number of years, Rahul is turning and T

_{1}= 100. Then the series can be re written as (T_{0}= 0)T

_{1}+ n × T_{0},T

_{2}= T_{1}+ n × 100,T3 = T

_{2}+ n × 100 = T_{1}+ (n-1) × 100 + n × 100T4 = T3 + n × 100 = T

_{1}+ (n - 2) × 100 + (n - 1) × 100 + n × 100....

T15 = T

_{14}+ 15 × 100= 1 × 100 + 2 × 100 + 3 × 100 + ........+ 15 × 100

All the terms after T

_{1}are in AP where, nth term is sum of n terms of the AP wherea = 100 and d = 100 n = 15

Hence T

_{15}== n(2a + (n-1)d)/2

= 15(200 + 14 × 100)/2

= 15 × 800

= 12000

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