Q1. What is the 12th term of the arithmetic sequence where the first term is 7 and the common difference is 5?
Solution:
Using the formula a_n = a + (n-1)d, we have a_12 = 7 + (12-1) * 5 = 7 + 55 = 62.
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Q2. Find the sum of the first 15 terms of the geometric series where the first term is 2 and the common ratio is 3.
Solution:
The sum of the first n terms of a geometric series is S_n = a(1 - r^n) / (1 - r). Here, a = 2, r = 3, n = 15. So, S_15 = 2(1 - 3^15) / (1 - 3) = 2(1 - 14348907) / -2 = 14348906.
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Q3. Find the sum of the first 5 terms of the series 1, 4, 9, 16, ...
Solution:
The series is the sum of squares: 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55.
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Q4. What is the 10th term of the arithmetic sequence where the first term is 5 and the common difference is 3?
Solution:
The nth term of an arithmetic sequence is given by a_n = a + (n-1)d. Here, a = 5, d = 3, n = 10. So, a_10 = 5 + (10-1) * 3 = 5 + 27 = 32.
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Q5. If the first term of a geometric series is 4 and the common ratio is 2, what is the 7th term?
Solution:
The nth term of a geometric series is given by a_n = ar^(n-1). Here, a = 4, r = 2, n = 7. So, a_7 = 4 * 2^(7-1) = 4 * 64 = 256.
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Q6. In a sequence where the nth term is given by n^2 + 2n, what is the 6th term?
Solution:
Substituting n = 6 into the formula: a_6 = 6^2 + 2(6) = 36 + 12 = 48.
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Q7. What is the 8th term of the sequence defined by a_n = 3n - 2?
Solution:
Substituting n = 8 into the formula: a_8 = 3(8) - 2 = 24 - 2 = 22.
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Q8. If the first term of an arithmetic series is 12 and the last term is 48, what is the common difference if there are 10 terms?
Solution:
In an arithmetic series, the last term can be expressed as a + (n-1)d. Here, 48 = 12 + (10-1)d. Thus, 48 - 12 = 9d, giving d = 4.
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Q9. What is the 5th term of the sequence defined by a_n = 2n^2 + 3n - 1?
Solution:
To find the 5th term, substitute n = 5 into the formula: a_5 = 2(5^2) + 3(5) - 1 = 2(25) + 15 - 1 = 50 + 15 - 1 = 64.
