The sum of $n$ terms of the series $1 + 1.5 +2 + 2.5+ 3 + ...$ is?

$\frac{n(n+3)}{4}$

The correct answer is Option (1) → $\frac{n(n+3)}{4}$

The given series is an Arithmetic Progression (A.P.):

$1,\; 1.5,\; 2,\; 2.5,\; 3,\; \ldots$

• First term a = 1
• Common difference d = 0.5

The nth term:

$a_n = a + (n-1)d = 1 + \frac{n-1}{2} = \frac{n+1}{2}$

Sum of first n terms:

$S_n = \frac{n}{2} (a + a_n)$

$S_n = \frac{n}{2} \left(1 + \frac{n+1}{2}\right) = \frac{n}{2} \cdot \frac{n+3}{2} = \frac{n(n+3)}{4}$