SOLUTION: An arithmetic sequence has first term a and common difference d. The sum of the first 25 terms is 15 times the sum of the first 4 terms. a) Find a in terms of d. b) Find the 55t

Algebra ->  Test -> SOLUTION: An arithmetic sequence has first term a and common difference d. The sum of the first 25 terms is 15 times the sum of the first 4 terms. a) Find a in terms of d. b) Find the 55t      Log On


   

Question 1186962: An arithmetic sequence has first term a and common difference d. The sum of the first 25
terms is 15 times the sum of the first 4 terms.
a) Find a in terms of d.
b) Find the 55th term in terms of a.
Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


a)

S(4) = (a)+(a+d)+(a+2d)+(a+3d) = 4a+(1+2+3)d = 4a+6d

S(25) = (a)+(a+d)+(a+2d)+...+(a+23d)+(a+24d) = 25a+(1+2+...+23+24)d = 25a+300d

S(25) is 15 times S(4):

25a%2B300d=15%284a%2B6d%29
25a%2B300d=60a%2B90d
210d=35a
a=210d%2F35
a=6d

b) t%2855%29=a%2B54d+=+6d%2B54d+=+60d+=+10a

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

a) Find a in terms of d


Sum of n+terms :
S%5Bn%5D=%28n%2F2%29%282a%2B%28n-1%29d%29
For 25 terms
S%5B25%5D=%2825%2F2%29%282a%2B%2825-1%29d%29
S%5B25%5D+=+25a+%2B+300d

For 4 terms
S%5B4%5D=%284%2F2%29%282a%2B%284-1%29d%29
S%5B4%5D+=+2%282a%2B3d%29
S%5B4%5D+=+4a+%2B+6d
given that the sum of the first 25 terms is 15+times the sum of the first 4 terms, so e have
25a+%2B+300d+=+15%284a+%2B+6d%29
25a+%2B+300d+=+60a+%2B+90d
300d-+90d+=+60a+-25a+
35a+=++210d
a+=++210d%2F35
a+=++6d

b) Find the 55th term in terms of a
a+=++6d=> d=++a%2F6
a%5B55%5D=+a%2B%2855-1%29%28a%2F6%29
a%5B55%5D=a%2B54a%2F6
a%5B55%5D=a%2B9a
a%5B55%5D=10a