## Answers

1st and 2nd term of an a.p = 4

a =1st term

a + d= 2nd term

a+ a +d =4

therefore, 2a+d=4

the 10th term = 19

a+9d=19

using simultaneous equation, by the elimination method

2a+d=4(eqn 1)

a+9d=19(eqn 2)

9(2a+d=4)

1(a+9d)=19

=18a+9d=36

-a+9d=19

=17a+0=17

=17a=17

=17 divided by 17=1

therefore a= 1

substitute for a in eqn 2

1+9d=19

9d=19-1=18

18divided by 9 =2

then 5th term= a+4d

6th term= a+5d

sub for a and d

a+4d= 1+4(2)=9

a+5d= 1+5(2)=11

11+9= 20