Meeting point — variable speed walker vs constant speed walker: Akbar starts from Agra towards Banaras at a constant 8 km/h. Birbal starts from Banaras towards Agra and walks 4 km in the first hour, 5 km in the second, 6 km in the third, and so on (increasing by 1 km each hour). The cities are 144 km apart. Where do they meet?

Introduction / Context:
This problem blends arithmetic progression (AP) distance accumulation with a constant-speed counterpart to determine a meeting location along a straight line.

Given Data / Assumptions:

• Distance Agra–Banaras = 144 km
• Akbar: constant speed 8 km/h
• Birbal: hour-1 = 4 km, hour-2 = 5 km, hour-3 = 6 km, … an AP with a = 4, d = 1
• Both start simultaneously toward each other.

Concept / Approach:
If they meet after n hours, Akbar covers 8n. Birbal covers S_n of the AP: S_n = n/2 * (2a + (n−1)d) = n/2 * (8 + n−1) = n/2 * (n + 7). The sum of distances equals 144 km.

Step-by-Step Solution:

Verification / Alternative check:
Symmetric split (72 + 72) equals 144 km, so they meet exactly in the middle.

Why Other Options Are Wrong:

• “In 6 h”, “In 8 h”: Do not satisfy the AP equation to sum 144 km.
• “80 km away from Banaras”: Incorrect location given computed 72 km from either end.
• “In 9 h”: The time is 9 h, but the question asks where they meet; the correct place is midway.

Common Pitfalls:
Confusing distance-per-hour AP with speed AP; here the hourly distances form an AP, not Birbal’s speed in km/h (though numerically identical here hour by hour).

Final Answer:
Midway between Agra and Banaras