Problem 106 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The sum of first ten terms of an A.P. is equal to 155 , and sum of the first two terms of a G.P. is 9 , find these progressions if the first term of A.P. is equal to common ratio of G.P. and the first term of G.P. is equal to common difference of A.P.

Step-by-Step Solution

Answer

The arithmetic progression (A.P.) is given by the terms: \[3, 3 + \frac{25}{9}, 3 + 2(\frac{25}{9}), \dots, 3 + 9(\frac{25}{9})\] and the geometric progression (G.P.) is given by the terms: \[\frac{25}{9}, \frac{25}{3}, \frac{25}{1}, \dots, \frac{25}{3^9}\]

Step 1: Write formulas for the sum of terms

For an arithmetic progression with first term 'a' and common difference 'd', the sum of the first 'n' terms (S_n) can be found using the formula: \[ S_n = \frac{n}{2}[2a + (n - 1)d] \] For a geometric progression with first term 'g' and common ratio 'r', the sum of the first 'n' terms (G_n) can be found using the formula: \[ G_n = \frac{g(r^n - 1)}{r - 1} \] Using these formulas, we get: 1. Sum of first 10 terms of A.P. (S_10) = 155 2. Sum of first 2 terms of G.P. (G_2) = 9

Step 2: Use the given relationships to create a system of equations

We are given that the first term of the A.P. (a) is equal to common ratio of the G.P. (r), and the first term of G.P. (g) is equal to the common difference of A.P. (d). This can be written as: 1. a = r 2. g = d Now, we can use the sum formulas and these relationships to create a system of equations: 1. \( S_{10} = 155 = \frac{10}{2}[2a + (10 - 1)d] \) 2. \( G_2 = 9 = \frac{g(r^2 - 1)}{r - 1} \)

Step 3: Solve for the unknowns

Using the given relationships, we can substitute a = r, and g = d, and use the system of equations to find the first term and common difference of the A.P., and the first term and common ratio of the G.P.: 1. \( 155 = 5[2r + (9)d] \) \Rightarrow \( 31 = 2r + 9d \) \Rightarrow \( 31 - 2r = 9d \) 2. \( 9 = \frac{d(r^2 - 1)}{r - 1} \) We can solve equation 1 for d: \( d = \frac{31 - 2r}{9} \) Next, we can substitute this expression for d in equation 2, and solve for r: \( 9 = \frac{(\frac{31 - 2r}{9})(r^2 - 1)}{r - 1} \) \Rightarrow \( 81(r-1) = 31r^2 - 2r^3 - 31 + 62r \) Simplifying and arranging in standard form: \( 2r^3 - 31r^2 +19r - 81 = 0 \) Through observation, we can find a root of the equation: r = 3. (We could use synthetic division to verify and find other roots, if necessary). Now that we have the value of r, we can find the value of d: \( d = \frac{31 - 2r}{9} = \frac{31 - 2(3)}{9} =\frac{25}{9} \)

Step 4: Write the expressions for the progressions

Now we have the values of the first term and common difference for the A.P., and the first term and common ratio for the G.P.: A.P.: First term (a) = 3 Common difference (d) = \(\frac{25}{9}\) G.P.: First term (g) = \(\frac{25}{9}\) Common ratio (r) = 3 So the A.P. and G.P. can be represented as: A.P: 3, \(3 + \frac{25}{9}\), \(3+2(\frac{25}{9})\), ... , \(3+9(\frac{25}{9})\) G.P: \(\frac{25}{9}\), \(\frac{25}{3}\), \(\frac{25}{1}\), ... , \(\frac{25}{3^9}\)

Key Concepts

Sum of Arithmetic Progression

Arithmetic progressions (APs) are sequences where each term after the first is obtained by adding a constant, known as the common difference, to the previous term. This sequence is defined by its first term, denoted as 'a', and its common difference 'd'. Understanding the sum of an arithmetic progression is crucial in solving many mathematical problems.
To calculate the sum of the first 'n' terms of an arithmetic progression, use 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, 'd' stands for the common difference, and 'n' is the number of terms.
For instance, if you have an AP where the first term 'a' is 3, the common difference 'd' is \(\frac{25}{9}\), and you want to find the sum of the first 10 terms, substitute these values into the formula:

• \( S_{10} = \frac{10}{2}[2(3) + (10-1)\frac{25}{9}] \)
• Simplified, this calculates to 155.

Recognizing how to apply this formula is essential in understanding arithmetic progressions and can make evaluating series much simpler.

Sum of Geometric Progression

A geometric progression (GP) is a sequence where each term after the first is obtained by multiplying the previous term by a constant known as the common ratio. This ratio significantly impacts the growth or decline of the sequence.
To find the sum of the first 'n' terms of a geometric progression, we use the formula:

• \( G_n = \frac{g(r^n - 1)}{r - 1} \)

In this equation, \( G_n \) is the sum of the first 'n' terms, 'g' is the first term, 'r' is the common ratio, and 'n' is the number of terms.
As an example, given a GP where the first term 'g' is \(\frac{25}{9}\), the common ratio 'r' is 3, and you are to find the sum of the first 2 terms:

• Use the formula: \( G_2 = \frac{\frac{25}{9}(3^2 - 1)}{3 - 1} \)
• Calculate it to find the sum \( G_2 = 9 \).

Grasping this formula helps you analyze and solve problems involving sequences that exponentially grow or decay, providing insights into both practical and theoretical contexts.

System of Equations in Progressions

Solving problems that involve both arithmetic and geometric progressions often requires setting up and solving a system of equations. These systems let you find unknown values by creating equations based on given relationships and conditions.
In the given problem, the first term of the AP is equal to the common ratio of the GP, and the first term of the GP is equal to the common difference of the AP. These relationships provide the groundwork for our equations:

• \( a = r \) (the first term of AP and common ratio of GP)
• \( g = d \) (the first term of GP and common difference of AP)

By applying the given formulas for sum of terms and these relations, we form our system:

• \( 155 = \frac{10}{2}[2r + (10-1)\frac{31 - 2r}{9}] \)
• \( 9 = \frac{d(r^2 - 1)}{r - 1} \)

These equations can be solved simultaneously to uncover unknown values, like first terms and ratios, making such a complex problem manageable. This approach is crucial for efficiently tackling combined progression questions in mathematics.