Question:
\(\frac{\text{2+3+5+…. to n terms}}{\text{2+5+8+….8 \,to \,8 \,\,terns}}\)=9, then the value of n is
\(\frac{\text{2+3+5+…. to n terms}}{\text{2+5+8+….8 \,to \,8 \,\,terns}}\)=9, then the value of n is
Updated On: Sep 3, 2024
- 20
- 40
- 10
- 30
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The Correct Option is D
Solution and Explanation
The correct option is (D): 30
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