Question

In an A.P. first term is $\frac{10}{3}$ and first 30 terms are non-negative such that sum of first 30 terms = $(T_{30})^3$, then d is equal to (where $T_n$ is $n^{\text{th}}$ term of A.P., and d is the common difference of A.P.)

• A. $\frac{3}{87}$
• B. $\frac{5}{87}$
• C. $\frac{7}{87}$
• D. $\frac{9}{87}$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In an A.P., $S_n = \frac{n}{2}(2a + (n-1)d)$ and $T_n = a + (n-1)d$.
Step 2: Key Formula or Approach:
Set up the equation $S_{30} = (T_{30})^3$ using $a = 10/3$.
Step 3: Detailed Explanation:
1. Sum $S_{30}$:
$S_{30} = \frac{30}{2}(2(10/3) + 29d) = 15(20/3 + 29d) = 100 + 435d$.
2. Term $T_{30}$:
$T_{30} = 10/3 + 29d$.
3. Let $29d = x$. Then:
$100 + 15x = (10/3 + x)^3$.
Expanding the right side is tedious; let's test $T_{30}$ as an integer or simple fraction.
Try $T_{30} = 5$:
$5^3 = 125$.
LHS: $100 + 15(5 - 10/3) = 100 + 15(5/3) = 100 + 25 = 125$.
Matches. So $T_{30} = 5$.
4. Solve for $d$:
$10/3 + 29d = 5 \implies 29d = 5 - 10/3 = 5/3$.
$d = \frac{5}{3 \cdot 29} = \frac{5}{87}$.
Step 4: Final Answer:
The common difference $d$ is $\frac{5}{87}$.