Sequences and Series Ex 9.3: Mastering Geometric Progressions (G.P.)

Key Definitions for Geometric Progression (G.P.)

Geometric Progression (G.P.)

A sequence of non-zero numbers is called a Geometric Progression if the ratio of any term to its preceding term is always constant. This constant is called the common ratio.

Common Ratio (r)

The constant ratio between consecutive terms in a G.P. It is calculated as r = a_k / a_{k-1}, where a_k is the current term and a_{k-1} is the previous term.

nth Term of a G.P. (a_n)

The formula to find the nth term of a G.P. is a_n = a * r^(n-1), where 'a' is the first term, 'r' is the common ratio, and 'n' is the term number.

Sum of n Terms of a G.P. (S_n)

The sum of the first 'n' terms is given by S_n = a(r^n - 1) / (r - 1) or S_n = a(1 - r^n) / (1 - r), provided r ≠ 1. The first form is convenient for |r| > 1 and the second for |r| < 1.

Understanding the Core G.P. Formulas

Let's dive deeper into the two most important formulas for G.P. that are the backbone of Exercise 9.3.

**1. The nth Term Formula: a_n = a * r^(n-1)**
Think of 'a' as your starting point. To get to the second term (a_2), you multiply by 'r' once (ar). To get to the third term (a_3), you multiply by 'r' twice (ar*r = ar²). Notice the pattern? To get to the nth term, you start with 'a' and perform 'n-1' multiplications by 'r'. This gives us the elegant formula a_n = ar^(n-1). It's a direct way to find any term without having to list out the entire sequence.

2. The Sum Formula: S_n = a(r^n - 1) / (r - 1)
The proof for this is insightful. Let S_n = a + ar + ar² + ... + ar^(n-1). Now, multiply this whole equation by 'r': rS_n = ar + ar² + ar³ + ... + ar^n. If you subtract the first equation from the second (rS_n - S_n), most terms cancel out, leaving you with S_n(r - 1) = ar^n - a. Rearranging this gives the formula. Why are there two versions? The second version, S_n = a(1 - r^n) / (1 - r), is simply the first one multiplied by -1 in both the numerator and denominator. We use it when |r| < 1 to keep our calculations with positive denominators, making life easier and reducing errors.

Worked Examples (NCERT Ex 9.3 Style)

• Problem 1: Find the 10th and nth terms of the G.P. 5, 25, 125, ... Step 1: Identify the first term (a) and the common ratio (r). The first term is a = 5. To find the common ratio, divide the second term by the first: r = 25 / 5 = 5. (Check with the next pair: 125 / 25 = 5. It's a G.P.) Step 2: Find the 10th term (a_10) using a_n = ar^(n-1). Here, n = 10. a_10 = 5 (5)^(10-1) = 5 5^9 = 5^10. Step 3: Find the nth term (a_n) using the same formula. a_n = 5 * (5)^(n-1) = 5^(1+n-1) = 5^n. Final Answer: The 10th term is 5^10 and the nth term is 5^n.
• Problem 2: In a G.P., the 3rd term is 24 and the 6th term is 192. Find the 10th term. Step 1: Set up equations using the nth term formula, a_n = ar^(n-1). 3rd term (a_3) = ar^(3-1) = ar² = 24 ---(i) 6th term (a_6) = ar^(6-1) = ar⁵ = 192 ---(ii) Step 2: Solve for 'r' by dividing equation (ii) by equation (i). (ar⁵) / (ar²) = 192 / 24 r³ = 8 r = 2 Step 3: Substitute 'r' back into equation (i) to find 'a'. a (2)² = 24 a 4 = 24 a = 6 Step 4: Now that we have a = 6 and r = 2, find the 10th term (a_10). a_10 = ar^(10-1) = ar⁹ = 6 (2)⁹ a_10 = 6 512 = 3072. Final Answer: The 10th term is 3072.

Exam Traps in Geometric Progression Problems

Be careful! G.P. problems have a few common pitfalls. Here's what to watch out for:

• Confusing G.P. and A.P.: The most common mistake. Always do a quick check. Is there a common ratio (G.P.) or a common difference (A.P.)? In an exam, write down r = a_2 / a_1 first to confirm it's a G.P. before applying any formulas.
• Negative Common Ratio (r < 0): When 'r' is negative, the terms of the G.P. will alternate in sign (e.g., 5, -10, 20, -40, ...). Be extremely careful with the exponent in ar^(n-1). A negative base raised to an even power is positive, but to an odd power is negative. For instance, (-2)^4 = 16 but (-2)^5 = -32.
• Sum Formula Choice: While S_n = a(r^n - 1) / (r - 1) and S_n = a(1 - r^n) / (1 - r) are mathematically identical (for r ≠ 1), choosing the right one simplifies your work. If r = 3, use the first formula to get 3-1=2 in the denominator. If r = 1/2, use the second one to get 1 - 1/2 = 1/2. This helps avoid negative signs in your denominator.

Practice Questions with Solutions

• Q: Find the sum of the first 7 terms of the G.P. 2, -6, 18, ... A: Step 1: Identify the first term 'a' and the common ratio 'r'. a = 2. r = (-6) / 2 = -3. Step 2: Since |r| > 1, it's convenient to use the sum formula S_n = a(r^n - 1) / (r - 1). Here n = 7. S_7 = 2 ((-3)^7 - 1) / (-3 - 1). Step 3: Calculate the values. (-3)^7 = -2187. S_7 = 2 (-2187 - 1) / (-4) = 2 * (-2188) / (-4). Step 4: Simplify the expression. S_7 = -4376 / -4 = 1094. Final answer: The sum of the first 7 terms is 1094.
• Q: How many terms of the G.P. 3, 3/2, 3/4, ... are needed to give a sum of 3069/512? A: Step 1: Identify a, r, and S_n. a = 3. r = (3/2) / 3 = 1/2. S_n = 3069/512. Step 2: Since |r| < 1, use the formula S_n = a(1 - r^n) / (1 - r). 3069/512 = 3 (1 - (1/2)^n) / (1 - 1/2). Step 3: Simplify and solve for n. 3069/512 = 3 (1 - (1/2)^n) / (1/2) = 6 (1 - (1/2)^n). Divide by 6: 3069 / (512 6) = 1 - (1/2)^n. 1023 / 1024 = 1 - (1/2)^n. (1/2)^n = 1 - 1023/1024 = 1/1024. Step 4: Express 1024 as a power of 2. 1024 = 2^10. (1/2)^n = 1 / 2^10 = (1/2)^10. Therefore, n = 10. Final answer: 10 terms are needed.
• Q: The sum of the first three terms of a G.P. is 16, and the sum of the next three terms is 128. Determine the first term and the common ratio. A: Step 1: Set up the equations based on the given information. Let the G.P. be a, ar, ar², ar³, ... Sum of first three terms: a + ar + ar² = a(1+r+r²) = 16 ---(i) Sum of next three terms: ar³ + ar⁴ + ar⁵ = ar³(1+r+r²) = 128 ---(ii) Step 2: Divide equation (ii) by equation (i) to find 'r'. [ar³(1+r+r²)] / [a(1+r+r²)] = 128 / 16. r³ = 8. r = 2. Step 3: Substitute r = 2 into equation (i) to find 'a'. a(1 + 2 + 2²) = 16. a(1 + 2 + 4) = 16. a(7) = 16. a = 16/7. Final answer: The first term 'a' is 16/7 and the common ratio 'r' is 2.
• Q: Find a G.P. for which the sum of the first two terms is -4 and the fifth term is 4 times the third term. A: Step 1: Translate the given statements into equations. Sum of first two terms: a + ar = -4 => a(1+r) = -4 ---(i) Fifth term is 4 times the third term: a₅ = 4 a₃. Step 2: Use the nth term formula on the second condition. ar⁴ = 4 ar². Divide both sides by ar² (assuming a≠0, r≠0): r² = 4. So, r = 2 or r = -2. Step 3: Case 1: If r = 2. Substitute into equation (i). a(1 + 2) = -4 => 3a = -4 => a = -4/3. The G.P. is -4/3, -8/3, -16/3, ... Step 4: Case 2: If r = -2. Substitute into equation (i). a(1 + (-2)) = -4 => a(-1) = -4 => a = 4. The G.P. is 4, -8, 16, ... Final answer: There are two possible G.P.s: (-4/3, -8/3, -16/3, ...) and (4, -8, 16, ...).

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