Savings in an arithmetic progression over 10 years: A man saves ₹ 1,45,000 in ten years. Each year after the first, he saves ₹ 2,000 more than in the previous year. How much did he save in the first year?

Introduction / Context:
This is a direct application of the sum formula for an arithmetic progression (AP) where yearly savings increase by a fixed amount. Identify the first term a and common difference d, then equate the AP sum to the total savings.

Given Data / Assumptions:

• Total years n = 10.
• Common difference d = ₹ 2000 per year.
• Total sum S = ₹ 1,45,000.

Concept / Approach:
For an AP, S = n/2 * [2a + (n − 1)d]. Solve for a given S, n, and d.

Step-by-Step Solution:
S = 10/2 * [2a + 9*2000] = 5 * (2a + 18000)145000 = 5 * (2a + 18000) ⇒ 2a + 18000 = 290002a = 11000 ⇒ a = 5500

Verification / Alternative check:
List a few terms: 5500, 7500, 9500, ... (increasing by 2000) and confirm that the AP sum equals ₹ 1,45,000.

Why Other Options Are Wrong:
They do not satisfy the AP sum equation with n = 10 and d = ₹ 2000.

Common Pitfalls:
Using n rather than (n − 1) with d; forgetting to halve the product when applying S = n/2 [...].

Final Answer:
₹ 5500