JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of the first \(40\) terms of the series. \(3+4+8+9+13+14+18+19+\ldots\) is \(( 102) \mathrm{m}\) then \(\mathrm{m}\) is equal to
- A \(20\)
- B \(5\)
- C \(10\)
- D \(25\)
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
Sum of the \(40\) terms of \(3+4+8+9+13+14+18+19 \ldots\) \(=(3+8+13+\ldots \text { upto } 20 \text { term })\)\(+[4+9+15+\ldots \text { upto } 20 \text { terms }]\) \(=10[\{6+19 \times 5\}+\{8+19 \times 5\}]\) \(=10 \times 204=20 \times 102\)
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