the common difference is d=5
So, the nth term is 4 + 5(n-1) or, 5n-1
how do I do this can any body show me how?
An arithmetic sequence begins
4, 9, 14, 19, 24, . . . .
(a) Find the common difference d for this sequence.
d =
(b) Find a formula for the nth term an of the sequence.
an =
(c) Find the 35th term of the sequence.
a35 =
4 answers
(a) What this means is, we are trying to find one number that is added or subtracted from each term to get the next term. Since this is an arithmetic sequence, the d will always be the same. You can find d by subtracting the first term from the second term (9-4) and getting d = 5
(b) nth term formula ...
(1st term) + d(n-1)
The nth term in this sequence is 4 + 5(n-1). You can check this by plugging the # of the term in for n. The 1st term would be 4 + 5(0) = 4
(c) plug 35 into our formula to get
4 + 5(35) = 4 + 175 = 179
(b) nth term formula ...
(1st term) + d(n-1)
The nth term in this sequence is 4 + 5(n-1). You can check this by plugging the # of the term in for n. The 1st term would be 4 + 5(0) = 4
(c) plug 35 into our formula to get
4 + 5(35) = 4 + 175 = 179
SORRY 35th term would be
4 + 5(35-1) = 4 + 5(34) = 4 + 170 = 174
4 + 5(35-1) = 4 + 5(34) = 4 + 170 = 174
THANKS NICK with your help i was able to solve my problem. Appreciate it!
For the arithmetic sequence beginning with the terms {9, 14, 19, 24, 29, 34 ...}, what is the sum of the first 23 terms
D=5
9+5(N-1)
9+5 (4-5)
9+5 (23-1)= 4+5(23) = 4+207
For the arithmetic sequence beginning with the terms {9, 14, 19, 24, 29, 34 ...}, what is the sum of the first 23 terms
D=5
9+5(N-1)
9+5 (4-5)
9+5 (23-1)= 4+5(23) = 4+207
1 answer