Question

find the term of the A.P 9,12,15 ... which is 39 more than its36th term

Answer 1

ap = a(n-1)d
a= first term d= difference
ap = 36 term
36th = 9(36-1)3
36th=9×35+3
36th=315+3
36th= 318

Answer 2

here ,a=9 and d=3
according to question,
a(n)+39=a(36)
9+(n-1)3+39=9+(36-1)3
(n-1)3=35×3-39
(n-1)3=105-39
(n-1)=66/3
n-1=22
n=23
answer
therefore,23th term of AP will be 39 more than 36th term