What is the 21st term in the sequence:?
回答 (3)
2020-10-09 11:39 pm
I use this equation as a general way of explaining the n'th term of an arithmetic sequence:
a(n) = a + b(n - 1)
where "a" is the first term (6)
and "b" is the common difference (4)
You want the 21st term (n), so we can substitute and solve for a(21):
a(n) = 6 + 4(n - 1)
a(21) = 6 + 4(21 - 1)
a(21) = 6 + 4(20)
a(21) = 6 + 80
a(21) = 86
a(n) = a + b(n - 1)
where "a" is the first term (6)
and "b" is the common difference (4)
You want the 21st term (n), so we can substitute and solve for a(21):
a(n) = 6 + 4(n - 1)
a(21) = 6 + 4(21 - 1)
a(21) = 6 + 4(20)
a(21) = 6 + 80
a(21) = 86
2020-10-09 11:30 pm
Sequence 6, 10, 14, 18,...is an AP of form a, a+d, a+2d,....,a+(n-1)d where am = 1st
term = t(1), d = common difference between consecutive terms;
ie., d = {t(n+1)-t(n), n = 1,2,3,......} and t(n) = a+(n-1)d.;
Here, a = 6 & d = 4;
Then t(n) = 6+(n-1)4;
For n = 21, we have t(21) = 6+(20)(4) = 86;
Choice D gives correct answer.
term = t(1), d = common difference between consecutive terms;
ie., d = {t(n+1)-t(n), n = 1,2,3,......} and t(n) = a+(n-1)d.;
Here, a = 6 & d = 4;
Then t(n) = 6+(n-1)4;
For n = 21, we have t(21) = 6+(20)(4) = 86;
Choice D gives correct answer.
2020-10-09 11:08 pm
the difference between two sequential terms is 4 so we can generalize this sequence as
S_n = 6 +4*(n-1) for n = 1, 2, 3, 4, ...
S_21 = 6+4*20 = 86
S_n = 6 +4*(n-1) for n = 1, 2, 3, 4, ...
S_21 = 6+4*20 = 86
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