Math Problem Statement

if the sum of the first three terms in a geometric progression is 112 and the sum of the next three terms is 14. how many terms does it take for the sum to be greater than 125

Solution

We need to solve this step by step:

Step 1: Represent the first three terms in a geometric progression.

Let the first term be aa and the common ratio be rr. The first three terms are: a,ar,ar2a, ar, ar^2 The sum of these terms is given as: a + ar + ar^2 = 112 \tag{1}

Step 2: Represent the next three terms.

The next three terms are: ar3,ar4,ar5ar^3, ar^4, ar^5 The sum of these terms is given as: ar^3 + ar^4 + ar^5 = 14 \tag{2}

Step 3: Factorize both equations.

From equation (1): a(1+r+r2)=112a(1 + r + r^2) = 112

From equation (2): ar3(1+r+r2)=14ar^3(1 + r + r^2) = 14

Step 4: Divide equation (2) by equation (1).

ar3(1+r+r2)a(1+r+r2)=14112\frac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \frac{14}{112} r3=18r^3 = \frac{1}{8} r=12r = \frac{1}{2}

Step 5: Solve for aa using r=12r = \frac{1}{2}.

Substitute r=12r = \frac{1}{2} into equation (1): a(1+12+14)=112a \left(1 + \frac{1}{2} + \frac{1}{4}\right) = 112 a(44+24+14)=112a \left(\frac{4}{4} + \frac{2}{4} + \frac{1}{4}\right) = 112 a74=112a \cdot \frac{7}{4} = 112 a=64a = 64

Step 6: General formula for the sum of terms.

The sum of the first nn terms of a geometric progression is: Sn=a1rn1r,for r<1S_n = a \frac{1 - r^n}{1 - r}, \quad \text{for } r < 1 Substitute a=64a = 64 and r=12r = \frac{1}{2}: Sn=641(12)n112S_n = 64 \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} Sn=642(1(12)n)S_n = 64 \cdot 2 \cdot \left(1 - \left(\frac{1}{2}\right)^n\right) Sn=128(1(12)n)S_n = 128 \left(1 - \left(\frac{1}{2}\right)^n\right)

Step 7: Determine nn for Sn>125S_n > 125.

128(1(12)n)>125128 \left(1 - \left(\frac{1}{2}\right)^n\right) > 125 1(12)n>1251281 - \left(\frac{1}{2}\right)^n > \frac{125}{128} (12)n<1125128\left(\frac{1}{2}\right)^n < 1 - \frac{125}{128} (12)n<3128\left(\frac{1}{2}\right)^n < \frac{3}{128} Take the logarithm (base 2): nlog2(12)<log2(3128)n \log_2\left(\frac{1}{2}\right) < \log_2\left(\frac{3}{128}\right) n<log2(3)7-n < \log_2(3) - 7 n>7log2(3)n > 7 - \log_2(3) Using log2(3)1.585\log_2(3) \approx 1.585: n>71.585n > 7 - 1.585 n>5.415n > 5.415 Since nn must be an integer, n=6n = 6.

Final Answer:

It takes 6 terms for the sum to exceed 125.

Would you like a more detailed explanation or clarification on any step?
Here are some related questions to explore:

  1. How do logarithmic properties help solve exponential equations?
  2. What is the derivation of the sum formula for a geometric progression?
  3. How does the common ratio influence the growth or decay of a sequence?
  4. Can this result be verified using numerical methods or simulations?
  5. How does the choice of base for the logarithm affect the calculation?

Tip: Always verify calculations involving logarithms with precise values or a calculator for better accuracy!

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Math Problem Analysis

Mathematical Concepts

Geometric Progression
Logarithms
Sequences and Series

Formulas

Sum of the first n terms of a geometric progression: S_n = a(1 - r^n) / (1 - r) for |r| < 1
Logarithm property: log_b(a^n) = n * log_b(a)

Theorems

Properties of geometric progressions
Logarithmic rules for solving exponential equations

Suitable Grade Level

Grades 10-12