Question

The first term of an A.P. of 30 non-negative terms is \(\frac{10}{3}\). If the sum of this A.P. is the cube of its last term, then its common difference is:

• A. \(\frac{5}{87}\)
• B. \(\frac{25}{83}\)
• C. \(\frac{15}{29}\)
• D. \(\frac{5}{29}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For an Arithmetic Progression (A.P.) with \(n\) terms, first term \(a\), and last term \(l\), the sum of the series is given by \(S_n = \frac{n}{2}(a + l)\).
Step 2: Key Formula or Approach:
Given: \(n = 30\), \(a = \frac{10}{3}\).
Let the last term be \(l\). The sum of the A.P. is given as \(l^3\).
\[ S_{30} = \frac{30}{2}(a + l) = 15 \left( \frac{10}{3} + l \right) \] We equate this to \(l^3\):
\[ 15 \left( \frac{10}{3} + l \right) = l^3 \] Step 3: Detailed Explanation:
Simplify the equation:
\[ 50 + 15l = l^3 \implies l^3 - 15l - 50 = 0 \] By Rational Root Theorem, we test integer factors of 50. Let's try \(l = 5\):
\[ 5^3 - 15(5) - 50 = 125 - 75 - 50 = 0 \] So \(l = 5\) is a root. Factoring the polynomial:
\[ (l - 5)(l^2 + 5l + 10) = 0 \] For the quadratic factor \(l^2 + 5l + 10 = 0\), the discriminant is \(\Delta = 5^2 - 4(1)(10) = 25 - 40 = -15<0\). Thus, there are no other real roots.
So the last term is \(l = 5\).
The last term of an A.P. is given by \(l = a + (n - 1)d\). Substituting the values:
\[ 5 = \frac{10}{3} + 29d \] \[ 29d = 5 - \frac{10}{3} = \frac{15 - 10}{3} = \frac{5}{3} \] Step 4: Final Answer:
Solving for the common difference \(d\):
\[ d = \frac{5}{3 \times 29} = \frac{5}{87} \]