Problem 44 - A Complete Resource Book in Mathematics for JEE Main

Question

If the sum of first \(n\) terms of two A.P's are in the ratio \(3 n+8: 7 n+15\), then the ratio of their 12 th terms is (A) \(8: 7\) (B) \(7: 16\) (C) \(74: 169\) (D) \(13: 47\)

Step-by-Step Solution

Answer

The ratio of their 12th terms is C) \(74: 169\).

Step 1: Write the formula for the sum of the first n terms

The sum of the first \( n \) terms of an arithmetic progression (A.P.) is given by the formula \( S_n = \frac{n}{2} (2a + (n-1)d) \), where \( a \) is the first term and \( d \) is the common difference.

Step 2: Express given ratios using the sum formulas

Using the sum formulas for the sums of the respective A.Ps, we state that \( \frac{\frac{n}{2} (2a_1 + (n-1)d_1)}{\frac{n}{2} (2a_2 + (n-1)d_2)} = \frac{3n + 8}{7n + 15} \). This gives us \( \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n + 8}{7n + 15} \).

Step 3: Express the nth term formula

The nth term of an A.P. is given by \( a_n = a + (n-1)d \). Therefore, the 12th term is \( a_{12} = a + 11d \).

Step 4: Find the ratio of their 12th terms

We need to find \( \frac{a_1 + 11d_1}{a_2 + 11d_2} \). Since the sum formulas are equal to \( \frac{3n + 8}{7n + 15} \), we can substitute \( n = 12 \) into the given ratio, as both corresponding terms belong to the same arithmetic series.

Step 5: Simplify the expressions for the 12th terms

Plug \( n = 12 \) into \( \frac{2a_1 + 11d_1}{2a_2 + 11d_2} = \frac{3 \times 12 + 8}{7 \times 12 + 15} \). Simplify this ratio to \( \frac{44}{99} \) or \( \frac{4}{9} \).

Step 6: Solve for the desired ratio

The ratio of the 12th terms (\( a_1 + 11d_1 \) and \( a_2 + 11d_2 \)) is the same as the ratio of simplified sums, \( \frac{4}{9} \). Therefore, the ratio of their 12th terms matches an answer choice.

Key Concepts

Sum of n terms

In an arithmetic progression (A.P.), understanding the sum of the first \( n \) terms is crucial. This sum is calculated using the formula:

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

Here, \( S_n \) represents the sum of the first \( n \) terms, \( a \) is the first term of the sequence, \( n \) is the number of terms you are summing, and \( d \) is the common difference.
This formula helps us to find out how large the sum of an arithmetic sequence will be up to any number of terms. By splitting the formula, \( n/2 \) is the number of pairs of terms, and \((2a + (n-1)d)\) calculates the sum of each pair. This breakdown provides insight into how each piece contributes to the total sum.

nth term formula

To fully grasp and work with arithmetic progressions, the nth term formula is key. This formula is expressed as:

• \( a_n = a + (n-1)d \)

The nth term \( a_n \) is calculated by starting with the first term \( a \) and adding the product of \( (n-1) \) and the common difference \( d \).
This formula helps to find any term in the sequence without listing out all prior terms. For instance, if we need the 12th term, we simply substitute \( n = 12 \) into the formula, giving us \( a_{12} = a + 11d \). It offers a simple way to directly access individual terms, making it an essential tool for working with arithmetic sequences.

ratio of terms

Comparing terms in two different arithmetic progressions can often be done through ratios. In solving problems involving arithmetic sequences, understanding the ratio of terms is vital.
Consider an example where the ratio of the sums of first \( n \) terms of two A.P.'s is given as \( \frac{3n + 8}{7n + 15} \). From this, we can express:

• \( \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n + 8}{7n + 15} \)

This tells us the relative comparison between their summation up to any number \( n \).
For the specific case of finding the ratio of the 12th terms from these progressions, we can simplify the given ratio by substituting \( n = 12 \), resulting in \( \frac{4}{9} \). This calculation highlights how ratios provide a method to analyze the structure and relationships within and between arithmetic progressions.