Question

The number of terms in the series 101+99+97+.....+47 is-

• A. 25
• B. 28
• C. 30
• D. 20

Answer 1

Answer: (B)

28

Answer 2

This is an arithmetic progression (AP) with the first term $a = 101$, common difference $d = 99 - 101 = -2$, and the last term $l = 47$. Using the formula for the $n$-th term of an AP, $a_n = a + (n-1)d$, we have $47 = 101 + (n-1)(-2)$. Solving for $n$: $47 - 101 = -2(n-1) \implies -54 = -2(n-1) \implies 27 = n-1 \implies n = 28$.