Question

in an arithmetic sequence, the first term is 4, the fourth term is 19, and the nth term is 99. find the common difference and the number of terms n

Answer 1

Answer:

common difference=5

number of term(n)=20

Step-by-step explanation:

a=4, $T_{4} =19 ,  T_{n}=99$

but T_{4} =19= a+3d

a+3d=19

4+3d=19

3d=19-4

3d=15

d=15/3

d=5

$T_{n}=99$ =a+(n-1)d

99=a+(n-1)d

99=4+(n-1)*5

99=4+5n-5

99=5n-1

5n-1=99

5n=99+1

5n=100

n=100/5

n=20