Question

find the 12 tearm of an ap 3,5,7,9.... 201 from the end of the following arthematic progression​

Answer 1

Answer:

a=3

d=5-3=2

an=201

an =a+(n-1) d

201 =3+(n-1) 2

201-3 =(n-1) 2

198 |2 =n-1

99+1 =n

n=100

ending 12 tearms is 89tearm

so

a89 =a+88d

=3+88×2

=3+176

=179