Question:
What is the sum of the following series?
$-64, -66, -68, \ldots, -100$
What is the sum of the following series?
$-64, -66, -68, \ldots, -100$
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For arithmetic progressions, always check number of terms using $a_n = a + (n-1)d$ before applying the sum formula $S_n = \tfrac{n}{2}(a+l)$.
Updated On: Dec 9, 2025
- $-1458$
- $-1558$
- $-1568$
- $-1664$
- None of the above
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The Correct Option is B
Solution and Explanation
The given sequence is an arithmetic progression (A.P.).
Step 1: Identify first term, last term, and common difference.
First term: $a = -64$
Common difference: $d = -66 - (-64) = -2$
Last term: $l = -100$
Step 2: Find the number of terms $n$.
General term of an A.P.: \[ a_n = a + (n-1)d \] Substitute $a_n = -100$: \[ -100 = -64 + (n-1)(-2) \] \[ -100 = -64 - 2n + 2 \] \[ -100 = -62 - 2n \] \[ -38 = -2n \quad \Rightarrow \quad n = 19 \] Step 3: Apply sum of A.P. formula.
\[ S_n = \frac{n}{2}(a + l) \] \[ S_{19} = \frac{19}{2}(-64 + (-100)) \] \[ S_{19} = \frac{19}{2}(-164) \] \[ S_{19} = 19 \times (-82) = -1558 \] \[ \boxed{-1558} \]
Step 1: Identify first term, last term, and common difference.
First term: $a = -64$
Common difference: $d = -66 - (-64) = -2$
Last term: $l = -100$
Step 2: Find the number of terms $n$.
General term of an A.P.: \[ a_n = a + (n-1)d \] Substitute $a_n = -100$: \[ -100 = -64 + (n-1)(-2) \] \[ -100 = -64 - 2n + 2 \] \[ -100 = -62 - 2n \] \[ -38 = -2n \quad \Rightarrow \quad n = 19 \] Step 3: Apply sum of A.P. formula.
\[ S_n = \frac{n}{2}(a + l) \] \[ S_{19} = \frac{19}{2}(-64 + (-100)) \] \[ S_{19} = \frac{19}{2}(-164) \] \[ S_{19} = 19 \times (-82) = -1558 \] \[ \boxed{-1558} \]
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