The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
a6= 19 = a + (n – 1) d
=19 = a + 5d …(i)
A17 = 41 = a + (n – 1) d
= 41 = a + 16d …(ii)
Subtracting (i) from (ii), we get
22 = 11d
d = 2
Now substituting the value of d in (i): 19 = a + 10
= a = 9
A40= a + (40 – 1) 2
= 9 + 78
= 87