Discussion
A man saves Rs. 200 in the first month, Rs. 250 in the second, Rs. 300 in the third and so on. How much will he have saved after 2 years?
Answer: B
The savings form an Arithmetic Progression with the first term \(a = 200\) and common difference \(d = 50\). We need to find the sum for 2 years, which is \(n=24\) months.
The formula for the sum is \(S_n = \frac{n}{2}[2a + (n-1)d]\).
\(S_{24} = \frac{24}{2}[2(200) + (24-1)50] = 12[400 + (23)(50)] = 12[400 + 1150] = 12(1550) = 18600\).
He will have saved Rs. 18,600 after 2 years.
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