Answer :
Answer:
4.
Step-by-step explanation:
To find the number of terms in a geometric progression where the sum [tex] ( S_n) [/tex] is given, we need to use the formula for the sum of a geometric series. Let's break it down step-by-step:
1. Identify the first term and common ratio:
The given series is [tex] 0.5 - 0.1 + 0.02 - \ldots [/tex] .
- The first term (a) is 0.5.
- The common ratio (r) can be found by dividing the second term by the first term:
[tex] r = \dfrac{-0.1}{0.5} = -0.2 [/tex]
2. Sum of the first n terms of a geometric series:
The sum [tex] S_n [/tex] of the first n terms of a geometric series with common ratio r is given by:
[tex]S_n = a \dfrac{1 - r^n}{1 - r} [/tex]
Since we know [tex] S_n = 0.416 [/tex], [tex] a = 0.5 [/tex], and [tex]r = -0.2 [/tex], we can plug these values into the formula:
[tex] 0.416 = 0.5 \dfrac{1 - (-0.2)^n}{1 - (-0.2)} [/tex]
3. Simplify the equation:
[tex]0.416 = 0.5 \dfrac{1 - (-0.2)^n}{1 + 0.2} [/tex]
[tex] 0.416 = 0.5 \dfrac{1 - (-0.2)^n}{1.2} [/tex]
[tex] 0.416 = \dfrac{0.5}{1.2} (1 - (-0.2)^n) [/tex]
[tex] 0.416 = \dfrac{0.5}{1.2} (1 - (-0.2)^n) [/tex]
Simplify the fraction:
[tex] 0.416 = \dfrac{5}{12} (1 - (-0.2)^n) [/tex]
4. Isolate the exponential term:
Multiply both sides by [tex] \dfrac{12}{5} [/tex]:
[tex] 0.416 \times \dfrac{12}{5} = 1 - (-0.2)^n [/tex]
[tex] 0.9984 = 1 - (-0.2)^n [/tex]
[tex] (-0.2)^n = 1 - 0.9984 [/tex]
[tex] (-0.2)^n = 0.0016 [/tex]
5. Solve for n:
Take the natural logarithm of both sides:
[tex] \ln((-0.2)^n) = \ln(0.0016) [/tex]
Use the property of logarithms [tex][\ln(a^b) = b \ln(a) [/tex]:
[tex] n \ln(-0.2) = \ln(0.0016) [/tex]
Since [tex] \ln(-0.2) [/tex] is undefined for real numbers, we consider the absolute values and use the logarithm properties for negative base in terms of complex numbers. Here we'll use:
[tex] n \ln(0.2) = \ln(0.0016) [/tex]
Calculate the logarithms:
[tex] \ln(0.2) \approx -1.6094 [/tex]
[tex]\ln(0.0016) \approx -6.4378 [/tex]
Now solve for n:
[tex] n = \dfrac{\ln(0.0016)}{\ln(0.2)} [/tex]
[tex] n = \dfrac{-6.4378}{-1.6094} [/tex]
[tex]n \approx 4 [/tex]
So, the number of terms in the series is approximately 4.
