Question:
The sum of third and ninth term of an Arithmetic Progression is 8. Find the sum of the first 11 terms of the progression.
The sum of third and ninth term of an Arithmetic Progression is 8. Find the sum of the first 11 terms of the progression.
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In an AP, the sum of terms equidistant from the ends is constant.
Updated On: Jun 26, 2026
- 44
- 22
- 19
- 37
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Use the properties of an AP: $T_n = a + (n-1)d$ and $S_n = \frac{n}{2}(2a + (n-1)d)$.
Step 2: Analysis
$T_3 + T_9 = 8$
$(a + 2d) + (a + 8d) = 8$
$2a + 10d = 8$.
Step 3: Calculation
Sum of first 11 terms: $S_{11} = \frac{11}{2}(2a + 10d)$.
Substitute the value from
Step 2:
$S_{11} = \frac{11}{2}(8) = 11 \times 4 = 44$. Final Answer: (1)
Use the properties of an AP: $T_n = a + (n-1)d$ and $S_n = \frac{n}{2}(2a + (n-1)d)$.
Step 2: Analysis
$T_3 + T_9 = 8$
$(a + 2d) + (a + 8d) = 8$
$2a + 10d = 8$.
Step 3: Calculation
Sum of first 11 terms: $S_{11} = \frac{11}{2}(2a + 10d)$.
Substitute the value from
Step 2:
$S_{11} = \frac{11}{2}(8) = 11 \times 4 = 44$. Final Answer: (1)
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