Question

find tn for an arithmetic progression on where t7=15,t13=27, also find t20 ​

Answer 1

Answer:

t7=a+6d=15. equation 1

t13=a+12d=27. equation 2

by subtracting equation 1and 2

a+6d-(a+12d)=15-27

-6d=-12

d=2

substituting the value of d in equation 1

a+6(2)=15

a+12=15

a=3

tn=a+(n-1)d

=3+(n-1)2

=3+2n-2

=1+2n

t20=1+2(20)

=1+40

=41