Problem 112 - A Complete Resource Book in Mathematics for JEE Main

Question

The sum of of first ten terms of an A.P. is equal to 155 and the sum of first two terms of a G.P. is 9 . If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to the common difference of the A.P, then (A) first term of the G.P. is \(\frac{2}{3}, 3\) (B) first term of the A.P. is \(\frac{2}{3}, 3\) (C) Common ratio of the G.P. is \(\frac{25}{2}, 2\) (D) Common difference of the A.P is \(\frac{2}{3}, 3\)

Step-by-Step Solution

Answer

First term of G.P. is \(\frac{2}{3}, 3\) and common ratio is \(2\).

Step 1: Define the variables

Let the first term of the A.P. be \(a\) and the common difference of the A.P. be \(d\). Hence, the sum of the first 10 terms of the A.P. can be given by the formula \(S_n = \frac{n}{2} (2a + (n-1)d)\). For the given problem \(S_{10} = 155\) and \(n = 10\).

Step 2: Find the first term and common difference of the A.P.

Since the sum of the first 10 terms of the A.P. is 155, we have:\[S_{10} = \frac{10}{2} (2a + 9d) = 155\]Simplifying gives:\[5(2a + 9d) = 155\]\[2a + 9d = 31\]

Step 3: Define the variables for G.P.

Let the first term of the G.P. be \(b\) and the common ratio be \(r\). According to the problem, the sum of the first two terms of the G.P. is 9, so:\[b(1 + r) = 9\]

Step 4: Connect A.P. with G.P.

From the problem statement, the first term of the A.P. \(a\) is equal to the common ratio \(r\) of the G.P., and the common difference \(d\) of the A.P. is equal to the first term \(b\) of the G.P. Thus, \(a = r\) and \(d = b\).

Step 5: Substitute values and solve equations

Using \(a = r\) and \(d = b\) in the equations obtained from steps 2 and 3:\[2r + 9b = 31\]\[b(1 + r) = 9\]Substitute \(b = d\) from step 4 into the equation \(b(1 + r) = 9\), giving:\[d(1 + r) = 9\]

Step 6: Solve the equations simultaneously

We have two equations:1. \(2r + 9d = 31\)2. \(d(1 + r) = 9\)Substitute possible values to satisfy both equations:If \(d = 3\), then \(1 + r = 3\) gives \(r = 2\), solving \(2r + 9d = 31\) gives correct values.If \(d = \frac{2}{3}\), solving numerically satisfies both equations as well.

Step 7: Conclusion on possible solutions

If \(d = 3\), then \(r = 2\) and \(d = \frac{2}{3}\) is another viable pair. Each pair represents the first term \(b\) of the G.P, common difference of the A.P., and respective relations in A.P. and G.P according to conditions given.

Key Concepts

Geometric Progression

A Geometric Progression (G.P.) is a sequence of numbers where each term after the first is obtained by multiplying the previous one by a fixed, non-zero number called the common ratio. This is one of the fundamental concepts in sequences and series.

In the exercise example, understanding a Geometric Progression helps in determining relations between terms of sequences given certain conditions. It is stated that the first term of the Geometric Progression is equal to the common difference of the Arithmetic Progression, which is quite an engaging way to see how these two types of progressions can interact.

• The formula for the n-th term in a G.P. is given by: \( a_n = a_1 \times r^{(n-1)} \), where \( a_1 \) is the first term and \( r \) is the common ratio.
• The sum of the first n terms of a G.P. is calculated as: \( S_n = a_1 \frac{1-r^n}{1-r} \), if \( r eq 1 \).

In our exercise, the significance of these formulas lies in solving the equation \( b(1 + r) = 9 \), where \( b \) is the first term linked to the A.P.'s common difference. Understanding this relationship is key to understanding how G.P. contributes to solving the problem.

Common Ratio

The common ratio in a Geometric Progression is important as it defines the relationship between consecutive terms. It shows how each term increases or decreases in the sequence.

In a G.P., the common ratio is found by dividing any term by the previous term. For example, if the sequence is 2, 4, 8, ... the common ratio is \( r = \frac{4}{2} = 2 \). When applied consistently, this ratio can exponentially grow or shrink the series. In relation to the given exercise, this ratio is intricately linked with the Arithmetic Progression in such a way that the first term of the A.P. equals the common ratio of the G.P.

• It is calculated as \( r = \frac{a_{n}}{a_{n-1}} \).
• If \( r > 1 \), the sequence is increasing, while \( 0 < r < 1 \) implies a decreasing sequence.
• If \( r < 0 \), the sequence alternates between positive and negative values.

The problem gives us two possible values for the common ratio, helping to decide the relationship between two linked sequences - the Arithmetic and Geometric Progression. The equation \( r = 2 \) arises as one solution for the ratio, demanding comprehension of its implications.

Common Difference

The common difference is a defining feature of an Arithmetic Progression (A.P.). It is the fixed amount that each term increments by, creating a consistent linear pattern.

In the exercise, the link between the A.P. and G.P. is reinforced through this aspect, as the first term of the G.P. is stated to equal the A.P.'s common difference. This creates an intriguing bond between the arithmetic and geometric series.

• The common difference \( d \) is calculated by subtracting the previous term from any term, i.e., \( d = a_{n} - a_{n-1} \).
• This constant difference means that the nth term of an A.P. can be given by: \( a_n = a_1 + (n-1)d \).
• It ensures a linear progression of the sequence in contrast to the exponential growth seen in G.P.

In the formula \( 2a + 9d = 31 \), the common difference assumes a critical role. It not only affects the sum of terms in the A.P. but also aligns the terms with those in the G.P., reinforcing the complex interplay between these progressions.