Problem 67 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The ratio between the sum of \(n\) term of two A.P.'s is \(7 n+1: 4 n+27\). Find the ratio between their \(n\) th terms.

Step-by-Step Solution

Answer

The ratio between their $n$ th terms is \(\boxed{\frac{7n+1}{4n+27}}\).

Step 1: Write the Formula for Sum of n Terms for Both A.P.s

We have two A.P.s, so let's write the formula for the sum of n terms for both of them: For A.P.1: \(S_{1n} = \frac{n}{2} (2a_1 + (n-1)d_1)\) For A.P.2: \(S_{2n} = \frac{n}{2} (2a_2 + (n-1)d_2)\) We are given that the ratio between the sum of n terms of both A.P.s is \(7n+1: 4n+27\).

Step 2: Write the Given Ratio Using the Sum Formulas

We can write the given ratio using the sum formulas like this: \(\frac{S_{1n}}{S_{2n}} = \frac{7n+1}{4n+27}\) Now we will substitute the sum formulas for both A.P.s into this expression: \(\frac{\frac{n}{2} (2a_1 + (n-1)d_1)}{\frac{n}{2} (2a_2 + (n-1)d_2)} = \frac{7n+1}{4n+27}\)

Step 3: Simplify the Ratio and Solve for the Required Ratio

Since both numerator and denominator have a common factor of \(\frac{n}{2}\), we can cancel them out: \(\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n+1}{4n+27}\) We have to find the ratio between their nth terms, so we will use the formula for the nth term of an A.P. for both of them: For A.P.1: \(T_{1n} = a_1 + (n-1)d_1\) For A.P.2: \(T_{2n} = a_2 + (n-1)d_2\) Now, let's write the required ratio using the nth term formulas: \(\frac{T_{1n}}{T_{2n}} = \frac{a_1 + (n-1)d_1}{a_2 + (n-1)d_2}\) But we already have an expression for this ratio from Step 3: \(\frac{T_{1n}}{T_{2n}} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n+1}{4n+27}\) So, the ratio between their nth terms is \(\boxed{\frac{7n+1}{4n+27}}\).

Key Concepts

nth term of an A.P.

An arithmetic progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference, denoted by \(d\). The nth term of an A.P. is a very important concept. It helps in identifying any term in the sequence without listing all the terms. To find the nth term, we use the formula:

• \(T_{n} = a + (n-1)d\)

Here, \(a\) is the first term of the sequence, \(n\) is the position of the term in the sequence, and \(d\) is the common difference.
In this problem, we needed to find the ratio between the nth terms of two different A.P.s. We have specific formulas for both A.P.1 and A.P.2:

• A.P.1: \(T_{1n} = a_1 + (n-1)d_1\)
• A.P.2: \(T_{2n} = a_2 + (n-1)d_2\)

Using these expressions, we arrived at the ratio of the nth terms.

sum of n terms in A.P.

The sum of the first \(n\) terms of an arithmetic progression gives us insight into the aggregate value of a number sequence. To calculate this sum, we can use the formula:

• \(S_{n} = \frac{n}{2} \times [2a + (n-1)d]\)

In the formula above, \(n\) represents the number of terms we want to sum, \(a\) is the first term, and \(d\) is the common difference. This formula essentially finds the sum by considering the average of the sequence's first and last terms, multiplied by the total number of terms.
In the exercise, the sums of two different A.P.s were compared by their ratio. We initially used the sum formulas for both progressions (A.P.1 and A.P.2) to express this given ratio:

• \(S_{1n} = \frac{n}{2}(2a_1 + (n-1)d_1)\)
• \(S_{2n} = \frac{n}{2}(2a_2 + (n-1)d_2)\)

The problem statement provided the ratio as \(\frac{7n+1}{4n+27}\), which was essential for finding the nth term ratio.

ratio of sums in A.P.

The ratio of sums in an arithmetic progression is a measure of how the total of one sequence compares to another over a specified number of terms \(n\). This is especially useful when comparing different sequences to assess their relative sizes.
In our problem, we began with a given ratio that expressed the sum of \(n\) terms for two sequences as a function of \(n\):

• \(\frac{S_{1n}}{S_{2n}} = \frac{7n+1}{4n+27}\)

To simplify the investigation into the nth terms, we utilized the ratio of the sums to cancel out common factors and derive a relationship between the nth terms directly. By recognizing that both the numerator and denominator contained a common factor of \(\frac{n}{2}\), we simplified the equation to compare directly the nth terms of both sequences.
The ratio of sums allowed us to bypass direct calculation of terminal differences and instead focus on how the components of the A.P.s were related through their expressions for sums and nth terms.