Question

Let \( a_1, a_2, a_3, \dots \) be an A.P. If \( \frac{a_1 + a_2 + \dots + a_{10}}{a_1 + a_2 + \dots + a_p} = \frac{100}{p^2}, p \neq 10 \), then \( \frac{a_{11}}{a_{10}} \) is equal to :

• A. \( \frac{121}{100} \)
• B. \( \frac{21}{19} \)
• C. \( \frac{19}{21} \)
• D. \( \frac{100}{121} \)

Hint:

If the ratio of sums of two A.P.s (or same A.P. with different \( n \)) is \( \frac{S_n}{S_m} = \frac{n^2}{m^2} \), then the ratio of their \( n \)-th terms is \( \frac{a_n}{a_m} = \frac{2n-1}{2m-1} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Given the ratio of sums \( \frac{S_{10}}{S_p} = \frac{10^2}{p^2} \).
This implies \( S_n \) is proportional to \( n^2 \), which is a specific property of an A.P. starting with the first term \( a = d/2 \).
Step 2: Key Formula or Approach:
If \( S_n = k \cdot n^2 \), then the \( n \)-th term is given by \( a_n = S_n - S_{n-1} \).
Step 3: Detailed Explanation:
From the given condition, we can generalize \( S_n = C n^2 \).
Then, \( a_n = S_n - S_{n-1} = C n^2 - C(n-1)^2 \)
\[ a_n = C[n^2 - (n^2 - 2n + 1)] = C(2n - 1) \]
We need to find \( \frac{a_{11}}{a_{10}} \):
\[ \frac{a_{11}}{a_{10}} = \frac{C(2(11) - 1)}{C(2(10) - 1)} = \frac{22 - 1}{20 - 1} = \frac{21}{19} \]
Step 4: Final Answer:
The ratio \( \frac{a_{11}}{a_{10}} \) is \( \frac{21}{19} \).