The 10th term of an AP is 52 and 16th term is 82. Find the 32nd term and the general term.
Let a be the first term and d be the common difference of the given A.P. Let the A.P. be a1, a2, a3..... an.....
It is given that
a10 = 52 and a16 = 82
a+(10–1)d=52 and a+(16–1)d=82
a+9d=52 .... (i)
and,
a+15d=82 .... (ii)
Subtracting equation (ii) from equation (i), we get
–6d=–30
d=5
Putting d=5 in equation (i), we get
a+45=52
a=7
a32=a+(32–1)d=7+31×5=162
and,
an=a+(n–1)d=7(n–1)×5=5n+2.
Hence a32=162 and an=5n+2.