In an arithmetic sequence the sum of the first nineterms is 279 and the sum of the first twenty terms is 1280.Then,
(a) what is the 5th term of the sequence ?
(b) what is the 16th term of the sequence?
(c) write the sequence .
Let the sequence be as following:
a-4d, a-3d, a-2d, a-d, a, ....
9th term will be: a+4d
Sum of first 9 terms = 9a = 279
(a) 5th term is a = 279/9 = 31
(b) Sum of 20 terms = 9a + (a+5d)+(a+6d)......+(a+15d) = 1280
so, 110d = 1280-31*20 = 660
d = 660/110= 6
16th term = a+11d = 31+11*6=97
(c)Substitute a & d to get the sequence
a-4d, a-3d, a-2d, a-d, a, ....
9th term will be: a+4d
Sum of first 9 terms = 9a = 279
(a) 5th term is a = 279/9 = 31
(b) Sum of 20 terms = 9a + (a+5d)+(a+6d)......+(a+15d) = 1280
so, 110d = 1280-31*20 = 660
d = 660/110= 6
16th term = a+11d = 31+11*6=97
(c)Substitute a & d to get the sequence