# 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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