Q. 46

# A man saved Rs. 3

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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