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
 

  • 2
S9=9/2(2a+8 d)=279
S20=20/1(2a+19d)=1280
=9(a+4d)=279
=10(2a+19d)=1280
9a+36d=279
20a+190d=1280.....(2)
a+4d=31.....(1)
((1)*10)
20a+190d=1290.....(2)
20a+80d=620.......(1)
(2)-(1) 110d=660
d=660/110=6
a+4*7=31
a+24=31
a=31-24=7
5th term =7+4*6=31
16th term=7+15*6=97
sequence=7,13,19,25.....
  • 5
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