How many terms are there in 20, 25, 30, ......, 140?

In an arithmetic progression (AP), consecutive terms differ by a constant amount called the common difference. If a is the first term, d is the common difference, and l is the last (nth) term, the number of terms n satisfies l = a + (n − 1)d. This relation lets us find n directly whenever a, d, and l are known.

1. Identify the given quantities for this progression: first term a = 20, common difference d = 25 − 20 = 5, and last term l = 140.

2. Substitute these values into l = a + (n − 1)d: 140 = 20 + (n − 1) × 5.

3. Isolate (n − 1): (n − 1) = (140 − 20) / 5 = 120 / 5 = 24.

4. Solve for n: n = 24 + 1 = 25.

To verify, compute the 25th term directly: a + (25 − 1)d = 20 + 24 × 5 = 20 + 120 = 140, which matches the given last term, confirming the count.

Hence, the progression 20, 25, 30, ……, 140 has 25 terms.