Math Problem Statement

Given x, x + 4, 2x + 2 are the first three positive terms of a GP, calculate the sum of the first 6 terms. The sum of the first 4 terms of a GP is 60. Its third term exceeds the first term by 8. If the first term is greater than 1, calculate the sum of the first 6 terms.

Solution

The image contains two math problems involving geometric progressions (GP). I'll break down each one:

Question 1:

"Given x,x+4,2x+2x, x + 4, 2x + 2 are the first three positive terms of a GP, calculate the sum of the first 6 terms."

We are given three terms of a geometric progression (GP). In a GP, the ratio between consecutive terms is constant, so we can use that information to find the value of xx and then calculate the sum of the first 6 terms.

Step-by-Step Approach:

  1. Set up the GP ratio:
    For a GP, the ratio between the second and first term is the same as the ratio between the third and second term.
    So,
    x+4x=2x+2x+4\frac{x+4}{x} = \frac{2x+2}{x+4}
  2. Solve for xx: Cross-multiply and simplify to find xx.
  3. Find the common ratio: Once xx is determined, substitute it back to find the common ratio rr.
  4. Sum formula: The sum of the first nn terms of a GP is given by: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r} where aa is the first term, rr is the common ratio, and nn is the number of terms. Use this to find the sum of the first 6 terms.

Question 2:

"The sum of the first 4 terms of a GP is 60. Its third term exceeds the first term by 8. If the first term is greater than 1, calculate the sum of the first 6 terms."

Here, we need to calculate the sum of the first 6 terms, given:

  • The sum of the first 4 terms is 60.
  • The third term exceeds the first by 8.
  • The first term is greater than 1.

Step-by-Step Approach:

  1. Set up equations: Let the first term be aa and the common ratio be rr. The four terms are: a,ar,ar2,ar3a, ar, ar^2, ar^3 The sum of the first 4 terms is: a(1+r+r2+r3)=60a(1 + r + r^2 + r^3) = 60 The condition that the third term exceeds the first by 8 gives: ar2=a+8ar^2 = a + 8
  2. Solve for aa and rr: Use these two equations to solve for aa and rr.
  3. Find the sum of the first 6 terms: After determining aa and rr, use the formula for the sum of the first 6 terms of a GP.

Let me know if you would like me to solve these problems fully or provide any further clarifications.

Five Related Questions:

  1. How do we determine the common ratio in a geometric progression?
  2. What is the difference between arithmetic and geometric progressions?
  3. How do you find the sum of an infinite geometric series?
  4. What happens if the common ratio is negative in a geometric progression?
  5. How do we derive the formula for the sum of the first nn terms of a GP?

Tip:

Always cross-check if your sequence is a valid GP by ensuring the ratio between consecutive terms remains constant.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometric Progression (GP)
Ratios
Equations

Formulas

Common ratio formula: r = term_n / term_(n-1)
Sum of the first n terms of a GP: S_n = a(1 - r^n) / (1 - r)

Theorems

Properties of Geometric Progression

Suitable Grade Level

Grades 10-12