Question

the fourth and fifth terms of an arithmetic sequence are 13 and 15. find the 20th term

Answer 1

let first term be a and common difference be d.
a+3d=13
a+4d=15
subtracting second eqn from the first,
we get d=2.
substituting in 1st eqn,
we get a+6=13
a=7.
therefore 20th term is a+19d=7+38=45

Answer 2

t4= a+3d=13
t5=a+ 4d=15
solving d=2 and a =7
t20 = a +19d=7+19*2=7+38=45