Problem 44 - A Complete Resource Book in Mathematics for JEE Main 2017

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

(C) 74: 169

Step 1: Write the formula for sum of n terms in an A.P.

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 the given condition using the sum formula

It is given that the sum of the first \( n \) terms of two A.P.s are in the ratio \( 3n+8 : 7n+15 \). Let \( S_n^{(1)} \) and \( S_n^{(2)} \) be the sums of the first \( n \) terms of the first and second A.P., respectively. We have: \[ \frac{S_n^{(1)}}{S_n^{(2)}} = \frac{3n+8}{7n+15}. \]

Step 3: Substitute the sum formula into the ratio

Substituting the sum formula into the given ratio, we can write: \[ \frac{\frac{n}{2} (2a_1 + (n-1)d_1)}{\frac{n}{2} (2a_2 + (n-1)d_2)} = \frac{3n+8}{7n+15}. \] Simplifying gives: \[ \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n+8}{7n+15}. \]

Step 4: Consider the ratio for the 12th term

The \( n \)th term of an A.P. is given by \( T_n = a + (n-1)d \). We need to find the ratio of the 12th terms, so we set \( n = 12 \). Thus, \( T_{12}^{(1)} = a_1 + 11d_1 \) and \( T_{12}^{(2)} = a_2 + 11d_2 \).

Step 5: Calculate the ratio for the 12th terms

Applying \( n = 12 \) to the simplified sum ratio: \[ \frac{2a_1 + 11d_1}{2a_2 + 11d_2} = \frac{3 \times 12 + 8}{7 \times 12 + 15} = \frac{44}{99}. \] Simplifying \( \frac{44}{99} \), we get \( \frac{4}{9}. \) Therefore, the ratio of the 12th terms is \( 4:9 \).

Key Concepts

Sum of n terms of A.P.

In an Arithmetic Progression (A.P.), the sum of the first \( n \) terms, denoted as \( S_n \), is key to understanding how series accumulate value. To calculate this, we use the formula: \[S_n = \frac{n}{2} (2a + (n-1)d),\]where:

• \( a \) is the first term of the sequence,
• \( d \) is the common difference between consecutive terms, and
• \( n \) is the number of terms.

This formula helps us find the sum by essentially averaging the first and last term, then multiplying by the total number of terms. In the exercise, we are given a ratio of sums for two A.P.s and use the above formula to express this relationship in terms of their first terms and common differences. The ability to substitute this sum formula is critical when dealing with comparisons between two or more sequences.

nth term of A.P.

An Arithmetic Progression (A.P.) is defined by each term increasing by a constant value, known as the common difference. To find any specific term, particularly the \( n \)th term of the sequence, we use the formula: \[T_n = a + (n-1)d,\]where:

• \( T_n \) is the \( n \)th term,
• \( a \) is the first term,
• \( d \) is the common difference,
• \( n \) is the term number.

This equation calculates the \( nth \) term by adding the product of the number of terms minus one and the common difference to the first term. In the original exercise, we're particularly interested in the 12th term ratio between two A.P.s. Knowing how to determine individual terms is essential for such calculations, giving you precise control over exploring specific characteristics of sequences.

Common difference in A.P.

The common difference \( d \) in an Arithmetic Progression (A.P.) is a fundamental aspect that dictates how much each term differs from the previous one. It is constant throughout the sequence. To identify the common difference, consider two consecutive terms, say \( T_{n} \) and \( T_{n+1} \), where the difference is given by:\[d = T_{n+1} - T_{n}.\]Alternatively, if you know the first term \( a \) and any other term \( T_n \), you can express \( d \) using:\[d = \frac{T_n - a}{n-1}.\]In the problem provided, the common difference is part of the ratio calculations for sums and individual terms, such as the 12th term. The comparison of these elements depends heavily on understanding how terms scale based on this consistent differential value. Recognizing\( d \) helps to reconstruct the entire sequence from its parts or manipulate it according to your needs.