Which term of the arithmetic progression will be72 more than its 41th term?
a = 8, d = 6,
Last term = an
an= a + (n – 1) d
= 8 + (n – 1) 6
= 2 + 6n (i)
Let, 41st term, a41 = 8 + (41 – 1) 6
= 8 + 40 * 6 = 248
Now the term is 72 more than its 41st term
an = 72 + a41
= 72 + 248 = 320
Putting this value in (i), we get
320 = 2 + 6n
6n = 318
n = 53
Hence, 53rd term of the given A.P. is 72 more than its 41st term