Question
Download Solution PDFThe sum of the 9th term and 12th term of the series \(\frac{1}{7}, \frac{1}{11}, \frac{1}{15}\) ..... is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Series: 1/7, 1/11, 1/15, ...
Formula used:
This is an AP in the denominators: 7, 11, 15, ...
General term: Tn = 1 / [4n + 3]
Calculation:
9th term = 1 / (4×9 + 3) = 1 / 39
12th term = 1 / (4×12 + 3) = 1 / 51
Sum = 1/39 + 1/51
LCM of 39 and 51 = 663
1/39 = 17/663, 1/51 = 13/663
Sum = (17 + 13)/663 = 30/663
Simplify: 30 ÷ 3 = 10, 663 ÷ 3 = 221
∴ The sum is 10/221
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