Question

The sum of series, $(-100) + (-95) + (-90) + \dots + 110 + 115 + 120$, is:

• A. 0
• B. 220
• C. 340
• D. 450
• E. None of the above

Hint:

For AP sums, always calculate $n$ first using the last term formula, then apply $S_n = \frac{n}{2}(a+l)$. This saves time and avoids manual addition errors.

Answer 1

The Correct Option is D

The given sequence is: \[ -100, -95, -90, \dots, 110, 115, 120 \] Step 1: Identify sequence type.
This is an Arithmetic Progression (AP) with: \[ a = -100, \quad d = 5, \quad l = 120 \] where $a$ = first term, $d$ = common difference, $l$ = last term.

Step 2: Find number of terms ($n$).
The general formula of the $n$-th term of an AP is: \[ a_n = a + (n-1)d \] Setting $a_n = 120$: \[ 120 = -100 + (n-1)(5) \] \[ 120 + 100 = (n-1)(5) \] \[ 220 = 5(n-1) \] \[ n-1 = 44 \quad \Rightarrow \quad n = 45 \]

Step 3: Use sum formula.
Sum of an AP: \[ S_n = \frac{n}{2}(a + l) \] Substitute values: \[ S_{45} = \frac{45}{2}(-100 + 120) \] \[ S_{45} = \frac{45}{2}(20) \] \[ S_{45} = 45 \times 10 = 450 \]

Final Answer: \[ \boxed{\text{D. 450}} \]