Q. 46

A man saved Rs. 33000 in 10 months. In each month after the first, he saved Rs. 100 more than he did in the preceding month. How much did he save in the first month?

Answer :

Let the amount of money the man saved in first month = Rs. x


Now, the amount of money he saved in second month = Rs.(x + 100)


The amount of money he saved in third month = Rs.(x + + 100 + 100)


This will continue for 10 months.


We get a an AP as x , x + 100, x + 200,… up to 10 terms.


Here, first term = x


Common difference = d = 100


Number of terms = n = 10


Total amount of money saved by the man = x + (x + 100) + (x + 200) + … up to 10 terms. = Rs. 33000 (given)


Sum of 10 terms of the Arithmetic Series = 33000


S10 =33000


(10/2) × [2a + (10 - 1)d] =33000


(10/2) × [2(x) + 9(100)] =33000


5 × [2x + 900] =33000


2x + 900 =6600


2x = 6600 - 900


2x = 5700


x = 2850


Amount of money saved by the man in first month = Rs. 2850


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