Consider the 15th term of the following APs.(A) 2, 7, 12...(B) 7, 10,…

The correct answer is Option (1) → (D), (B), (A), (C)

In an arithmetic progression (AP) each term is the one before it plus a fixed common difference \(d\), so the \(n\)th term is \(a+(n-1)d\) and the 15th term is \(a+14d\). The last term 253 printed in (C) only tells where that AP stops; it does not change its 15th term.

• (A) first term 2, common difference 5: 15th term \(2+14\times5=72\)

• (B) first term 7, common difference \(10-7=3\): 15th term \(7+14\times3=49\)

• (C) first term 3, common difference 5: 15th term \(3+14\times5=73\)

• (D) first term 5, common difference 2: 15th term \(5+14\times2=33\)

Note: It seems that there is a mistake in the original NTA paper, which prints the fourth term of (B) as 17 when it should read 16. The first three terms 7, 10, 13 rise by 3 each time and the question calls every series an AP, so the next term is 16; the answer is not affected. A series that climbs by 3 each time suddenly takes a step of 4 in NTA’s paper.