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To find the sum of the first 10 terms of the arithmetic progression (AP) given that the 3³ term is 1 and the 5⁵ term is 7, we must first determine the common difference [tex]\(d\)[/tex] and the first term [tex]\(a\)[/tex] of the series. Here are the detailed steps:
1. Identify the given terms:
- The term at position [tex]\(3^3 = 27\)[/tex] is 1.
- The term at position [tex]\(5^5 = 3125\)[/tex] is 7.
2. Establish the general formula for the nth term in an AP:
[tex]\[ a_n = a + (n - 1) \cdot d \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(d\)[/tex] is the common difference.
3. Set up equations using the known terms:
- For the term at position 27:
[tex]\[ a + (27 - 1) \cdot d = 1 \][/tex]
Simplifies to:
[tex]\[ a + 26d = 1 \quad \text{(Equation 1)} \][/tex]
- For the term at position 3125:
[tex]\[ a + (3125 - 1) \cdot d = 7 \][/tex]
Simplifies to:
[tex]\[ a + 3124d = 7 \quad \text{(Equation 2)} \][/tex]
4. Solve the system of equations to find [tex]\(a\)[/tex] and [tex]\(d\)[/tex]:
Subtract Equation 1 from Equation 2:
[tex]\[ (a + 3124d) - (a + 26d) = 7 - 1 \][/tex]
[tex]\[ 3098d = 6 \][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{6}{3098} \approx 0.001936733376371853 \][/tex]
5. Substitute [tex]\(d\)[/tex] back into Equation 1 to find [tex]\(a\)[/tex]:
[tex]\[ a + 26 \cdot d = 1 \][/tex]
[tex]\[ a + 26 \cdot 0.001936733376371853 \approx 1 \][/tex]
[tex]\[ a + 0.050355067785668147 \approx 1 \][/tex]
[tex]\[ a \approx 0.9496449322143319 \][/tex]
6. Calculate the sum of the first 10 terms [tex]\(S_{10}\)[/tex]:
The formula for the sum of the first [tex]\(n\)[/tex] terms of an AP is:
[tex]\[ S_n = \frac{n}{2} (2a + (n - 1)d) \][/tex]
For [tex]\(n = 10\)[/tex]:
[tex]\[ S_{10} = \frac{10}{2} [2 \cdot 0.9496449322143319 + (10 - 1) \cdot 0.001936733376371853] \][/tex]
[tex]\[ S_{10} = 5 [1.8992898644286638 + 9 \cdot 0.001936733376371853] \][/tex]
[tex]\[ S_{10} = 5 [1.8992898644286638 + 0.017430600387346677] \][/tex]
[tex]\[ S_{10} = 5 \times 1.9167204648160105 \][/tex]
[tex]\[ S_{10} \approx 9.583602324080053 \][/tex]
Thus, the sum of the first 10 terms of the arithmetic progression is approximately [tex]\(9.583602324080053\)[/tex].
1. Identify the given terms:
- The term at position [tex]\(3^3 = 27\)[/tex] is 1.
- The term at position [tex]\(5^5 = 3125\)[/tex] is 7.
2. Establish the general formula for the nth term in an AP:
[tex]\[ a_n = a + (n - 1) \cdot d \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(d\)[/tex] is the common difference.
3. Set up equations using the known terms:
- For the term at position 27:
[tex]\[ a + (27 - 1) \cdot d = 1 \][/tex]
Simplifies to:
[tex]\[ a + 26d = 1 \quad \text{(Equation 1)} \][/tex]
- For the term at position 3125:
[tex]\[ a + (3125 - 1) \cdot d = 7 \][/tex]
Simplifies to:
[tex]\[ a + 3124d = 7 \quad \text{(Equation 2)} \][/tex]
4. Solve the system of equations to find [tex]\(a\)[/tex] and [tex]\(d\)[/tex]:
Subtract Equation 1 from Equation 2:
[tex]\[ (a + 3124d) - (a + 26d) = 7 - 1 \][/tex]
[tex]\[ 3098d = 6 \][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{6}{3098} \approx 0.001936733376371853 \][/tex]
5. Substitute [tex]\(d\)[/tex] back into Equation 1 to find [tex]\(a\)[/tex]:
[tex]\[ a + 26 \cdot d = 1 \][/tex]
[tex]\[ a + 26 \cdot 0.001936733376371853 \approx 1 \][/tex]
[tex]\[ a + 0.050355067785668147 \approx 1 \][/tex]
[tex]\[ a \approx 0.9496449322143319 \][/tex]
6. Calculate the sum of the first 10 terms [tex]\(S_{10}\)[/tex]:
The formula for the sum of the first [tex]\(n\)[/tex] terms of an AP is:
[tex]\[ S_n = \frac{n}{2} (2a + (n - 1)d) \][/tex]
For [tex]\(n = 10\)[/tex]:
[tex]\[ S_{10} = \frac{10}{2} [2 \cdot 0.9496449322143319 + (10 - 1) \cdot 0.001936733376371853] \][/tex]
[tex]\[ S_{10} = 5 [1.8992898644286638 + 9 \cdot 0.001936733376371853] \][/tex]
[tex]\[ S_{10} = 5 [1.8992898644286638 + 0.017430600387346677] \][/tex]
[tex]\[ S_{10} = 5 \times 1.9167204648160105 \][/tex]
[tex]\[ S_{10} \approx 9.583602324080053 \][/tex]
Thus, the sum of the first 10 terms of the arithmetic progression is approximately [tex]\(9.583602324080053\)[/tex].
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