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What is the sum of first six terms of a sequence whose nth term is 8 - n?
nth term is 8 - n. an = 8 - n, so the sequence is {7, 6, 5, 4, 3, 2,...} (this is a decreasing sequence since the successor term is smaller than the nth term). So, the sum of first six terms of the sequence is 27.
What are the first five terms of the sequence whose nth term is 5n-4?
9
Find the quadratic sequences nth term for these 4 sequences which are separated by the letter i iii 7 10 15 22 21 42 iii 2 9 18 29 42 57 iii 4 15 32 55 85 119 iii 5 12 27 50 81 120?
Check if the given sequences are quadratic sequences. 7 10 15 22 21 42 The first difference: 3 5 7 1 21. The second difference: 2 2 6 20. Since the second difference is not constant, then the given sequence is not a quadratic sequence. 2 9 18 29 42 57 The first difference: 7, 9, 11, 13, 15. The second difference: 2 2 2 2. Since the second difference is constant, then the given sequence is a quadratic sequence. Therefore, contains a n2 term. Let n = 1, 2, 3, 4, 5, 6, ... Now, let's refer the n2 terms as, 1, 4, 9, 16, 25, 36. As you see, the terms of the given sequence and n2 terms differ by 1, 5, 9, 13, 17, 21 which is an arithmetic sequence,say {an} with a common difference d = 4 and the first term a = 1. Thus, the nth term formula for this arithmetic sequence is an = a + (n - 1)d = 1 + 4(n - 1) = 4n - 3. Therefore, we can find any nth term of the given sequence by using the formula, nth term = n2 + 4n - 3 (check, for n = 1, 2, 3, 4, 5, 6, ... and you'll obtain the given sequence) 4 15 32 55 85 119 The first difference: 11, 17, 23, 30, 34. The second difference: 6 6 7 4. Since the second difference is not constant, then the given sequence is not a quadratic sequence. 5 12 27 50 81 120 The first difference: 7, 15, 23, 31, 39. The second difference: 8 8 8 8. Since the second difference is constant, then the given sequence is a quadratic sequence. I tried to refer the square terms of sequences such as n2, 2n2, 3n2, but they didn't work, because when I subtracted their terms from the terms of the original sequence I couldn't find a common difference among the terms of those resulted sequences. But, 4n2 works. Let n = 1, 2, 3, 4, 5, 6, ... Now, let's refer the 4n2 terms as, 4, 16, 36, 64, 100, 144. As you see, the terms of the given sequence and 4n2 terms differ by 1, -4, -9, -14, -19, -24 which is an arithmetic sequence, say {an} with a common difference d = -5 and the first term a = 1. Thus, the nth term formula for this arithmetic sequence is an = a + (n - 1)d = 1 -5(n - 1) = -5n + 6. Therefore, we can find any nth term of the given sequence by using the formula, nth term = 4n2 - 5n + 6 (check, for n = 1, 2, 3, 4, 5, 6, ... and you'll obtain the given sequence)
What is the common ratio of the geometric sequence 625 125 25 5 1?
It is 0.2
What is the sum of first six terms of a sequence whose nth term is 4n-2?
The first 6 terms would be when n is 1, 2, 3, 4, 5, and 6. And since the nth term is 3 - 4n, you would simply substitute the first six numbers for n:* n = 1; 3 - 4(1); 3 - 4; -1 * n = 2; 3 - 4(2); 3 - 8; -5 * n = 3; 3 - 4(3); 3 - 12; -9 * n = 4; 3 - 4(4); 3 - 16; -13 * n = 5; 3 - 4(5); 3 - 20; -17 * n = 6; 3 - 4(6); 3 - 24; -21 -1, -5, -9, -13, -17, -21
