JEE Mains · Maths · STD 11 - 8. sequence and series
The first term of an A.P. of \(30\) non-negative terms is \(\dfrac{10}{3}\). If the sum of this A.P. is the cube of its last term, then its common difference is:
- A \(\dfrac{5}{87}\)
- B \(\dfrac{25}{83}\)
- C \(\dfrac{15}{29}\)
- D \(\dfrac{5}{29}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{5}{87}\)
Step-by-step Solution
Detailed explanation
Given \(a = \dfrac{10}{3}\) and \(n = 30\). Let the last term be \(l\) and the common difference be \(d\). The sum of the A.P. is given by \(S_{30} = \dfrac{30}{2}(a + l) = 15\left(\dfrac{10}{3} + l\right)\). According to the given condition, \(S_{30} = l^3\).…
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