JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(A_1, A_2, A_3, \ldots, A_{39}\) be \(39\) arithmetic means between the numbers \(59\) and \(159\). Then the mean of \(A_{25}, A_{28}, A_{31}\) and \(A_{36}\) is equal to :
- A \(129\)
- B \(136\)
- C \(131.50\)
- D \(134\)
Answer & Solution
Correct Answer
(D) \(134\)
Step-by-step Solution
Detailed explanation
The sequence \(59, A_1, A_2, \ldots, A_{39}, 159\) forms an arithmetic progression with \(41\) terms. The common difference \(d\) is given by \(d = \dfrac{159 - 59}{39 + 1} = \dfrac{100}{40} = 2.5\) The \(k\)-th arithmetic mean is \(A_k = 59 + k \cdot d\) The required mean is…
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