Question

fine the 5th term of arithmetic progression whose 12 term exceeds the 5th term by 14 and the sum of both terms is 36.​

Answer 1

Answer:

12th term=a+11d

5th term=a+4d

acc to ques.

(a+11d)-(a+4d)=14

7d=14

d=2

also (a+11d)+(a+4d)=36

putting the value of d and solving for a we get

a=3

mth term=a+(m-1)d

=3+(m-1)2

mth term=1+2m