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

To solve this problem, we need to understand the properties of arithmetic progressions (A.P.) and how sums of terms work in an A.P.

An arithmetic progression is a sequence of numbers in which the difference of any two successive members is a constant. If \(a_1\) is the first term and \(d\) is the common difference, the \(n\)-th term is given by:

\(a_n = a_1 + (n-1) \cdot d\) 

The sum of the first \(n\) terms of an A.P. is given by:

\(S_n = \frac{n}{2} (2a_1 + (n-1)d)\)

Let's apply this to the problem:

1. The condition given is:
2. First, calculate \(S_{10}\):
3. Now, calculate \(S_p\):
4. Substitute these into the given equation:
5. By cross-multiplying, we get:
6. Simplifying the equation:
7. Expanding both sides:
8. Rearrange the equation:
9. Isolate terms:
10. Solve for \(p\):
11. Now, let's find \(\frac{a_{11}}{a_{10}}\):
12. Then, the ratio \(\frac{a_{11}}{a_{10}}\) is:
13. This simplifies to:
14. Given the specific values obtained earlier, and incorporating the derived conditions, it aligns with the choice of \(\frac{21}{19}\). Therefore, the correct ratio is:

The answer is \(\frac{21}{19}\).