Math Problem Statement

In a GP where terms are all positive, the third term exceeds the first term by 32 while the fifth term exceeds the first term by 320. Calculate the first term.

Solution

I can help extract the math problems from your image and solve them. Let me first transcribe the relevant parts of the image.


  1. Problem 6:
    In a GP where terms are all positive, the third term exceeds the first term by 32, while the fifth term exceeds the first term by 320. Calculate the first term.

  2. Problem 7:
    Find the geometric means of the following:
    (a) 3,123, 12
    (b) 4,364, 36
    (c) 5,20,455, 20, 45

  3. Problem 8:
    Find the indicated sums of these GPs:
    (i) 5,10,20,5, 10, 20, \dots find S6S_6 and S8S_8.
    (ii) 3,6,12,3, -6, 12, \dots find S5S_5 and S6S_6.

  4. Problem 9:
    Given a+ar+ar2+ar3=40a + a r + a r^2 + a r^3 = 40, then the first positive term of a GP, calculate the sum of the first 6 terms of the GP.

  5. Problem 11:
    The sum of the first 3 terms of a GP is 117, and the sum of the next 3 terms is 36. Calculate the third term.


I will begin solving these one by one. Let's start with Problem 6.


Problem 6:

Given that:

  • The third term exceeds the first term by 32, i.e., T3=a+32T_3 = a + 32.
  • The fifth term exceeds the first term by 320, i.e., T5=a+320T_5 = a + 320.

The general formula for the nn-th term of a geometric progression (GP) is: Tn=arn1T_n = a \cdot r^{n-1}

For the third term: T3=ar2=a+32T_3 = a \cdot r^2 = a + 32

For the fifth term: T5=ar4=a+320T_5 = a \cdot r^4 = a + 320

Now, I will solve this system of equations to find aa (the first term) and rr (the common ratio). Let's compute.The solutions for the first term aa and the common ratio rr are:

  • a=4a = 4, r=3r = -3
  • a=4a = 4, r=3r = 3

Since the terms of the GP are stated to be positive, we discard the solution with r=3r = -3. Therefore, the first term is a=4a = 4 and the common ratio is r=3r = 3.

Would you like to proceed with solving the next problem, or do you have any questions on this solution?

Here are some relative questions to expand on this problem:

  1. What is the formula for the nn-th term of a geometric progression?
  2. How does the common ratio affect the growth of a geometric progression?
  3. What is the sum of the first 5 terms of this GP?
  4. What happens to the sequence if the common ratio is negative?
  5. How do we find the sum of infinite terms in a geometric progression?

Tip: In a GP, if the common ratio is greater than 1, the terms grow rapidly.

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Math Problem Analysis

Mathematical Concepts

Geometric Progression
Exponential Growth

Formulas

T_n = a * r^(n-1)
T_3 = a * r^2 = a + 32
T_5 = a * r^4 = a + 320

Theorems

Formula for nth term of a geometric progression

Suitable Grade Level

Grades 10-12