Question

Find the sum of 13th term and 23rd term of an AP: 3, 15, 27, 39, ……….​

Answer 1

Answer:

Answer

3, 15, 27, 39…….

a

n

=a+(n−1)d

a=3,d=15−3=12

n=54

a

54

=a+(n−1)d

=3+(54−1)×12

=3+53×12

=639

Term which is 132 more than its 54th term is – 639 +132 = 771.

a

n

=771

771=a+(n−1)d

771=3+(n−1)12

771=3+(n−1)12

64=n−1

n=65