Question
The 11 term of an arithmetic sequence is 57 and the sum of the first and fourth terms is 29 determine the first three terms and the nth term
Answers
Answered by
Bot
First term: 8
Second term: 15
Third term: 22
nth term: an = 8 + (n - 1) × 7
Second term: 15
Third term: 22
nth term: an = 8 + (n - 1) × 7
Answered by
I Know
with no steps shown, only the incorrect answer, this reply by the bot is useless.
a+10d = 57
a + a + 3d = 29 ---> 2a + 3d = 29
double the first:
2a + 20d = 114
subtract the 2nd:
17d = 85
d = 5 , then a+50 = 57
a = 7
the terms are 7, 12, 17, 22, 27, ....
term(n) = a + (n-1)d
= 7 + (n-1)(5)
= 2 + 5n
a+10d = 57
a + a + 3d = 29 ---> 2a + 3d = 29
double the first:
2a + 20d = 114
subtract the 2nd:
17d = 85
d = 5 , then a+50 = 57
a = 7
the terms are 7, 12, 17, 22, 27, ....
term(n) = a + (n-1)d
= 7 + (n-1)(5)
= 2 + 5n