Consider the arithmetic sequence 1,5,9.... Can the difference between any two terms of this sequence be 101?
Answers
Step-by-step explanation:
Consider an arithmetic sequence whose 7th term is 34 and the 15th terms are 66. What is the common difference and what is the 20th term?
To answer this question, we will build two equations with two unknowns, using
a(n) = a(1) + (n-1)d
so,
34 = a(1) + 6d
66 = a(1) +14d
Using elimination by multiply the first term by -1, we get:
-34 = -a(1) -6d
66 = a(1) + 14d
which becomes,
32 = 8d
d = 4
plugging that difference into the first equation, we get
34 = a(1) +6(4)
a(1) = 10
Since we know now know the difference and the first term, we use the first equation is get the 20th term.
a(20) = 10 + (20–1)4
a(20) = 86.
So, the answers to this problem are 4 for the difference, and 86 for the 20th terms.
Step-by-step explanation:
No its a arithmetic progression there is common difference and it is 4 so there will not difference of 101
if you are asking about if any term will be 101 then lets find it
101 os not divisble by 4 so it is not term of given arithmetic sequence
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