If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.
a5=31
a +4d =31 (i)
a25= 140 + a5
a + 24d= 140 + 31
31 -4d +24d=171 [ from (i) ]
20d=140
d=7
Putting the value of d in (i)
a= 31 – 4(7) = 3
a1 = a=3
a2 =a +d=3 +7=10
a3 =a +2d=3 + 2(7) =17
Hence, the A.P. is 3, 10, 17,…