Question

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

Answer 1

Answer:

tn=1+2n

t20=41

Step-by-step explanation:

tn=a+(n-1)d

therefore

t7=a+(7-1)d=a+6d=15 equation 1

t13=a+(13-1)d=a+12d=27. equation 2

equation 1- equation 2

a+6d=15

-(a+12d=27)

therefore

-6d=-12

d=2

substituting the value of d in equation 1

therefore

a+6d=15

a+6(2)=15

a+12=15

a=3

therefore

tn=a+(n-1)d

= 3+(n-1)2

=3+2n-2

=1+2n

therefore

t20=1+2n

=1+2(20)

=1+40

=41

Hope it helps