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
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Answered by
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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
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
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