JEE Mains · Maths · STD 11 - 8. sequence and series
Given an \(A.P.\) whose terms are all positive integers. The sum of its first nine terms is greater than \(200\) and less than \(220\). If the second term in it is \(12\), then its \(4^{th}\) term is
- A \(8\)
- B \(16\)
- C \(20\)
- D \(24\)
Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
Let \(a\) be the frist term and \(d\) be the common difference of given \(A.P.\) Second term,\(a+d=12\) .....\((1)\) Sum of frist nine terms, \({S_9} = \frac{9}{2}\left( {2a + 8d} \right) = 9\left( {a + 4d} \right)\) Given that \({S_9}\) is more than \(200\) and less than…
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