determine the A.P whose fourth term is 18 and the difference of the ninth term from the fifteenth term is 30.

Answer 1

An arithmetic progression (A.P) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the nth term of the sequence is given by:

a n = a 1 + ( n - 1 ) d

a 1 = first term

In this case:

a 8 = a 1 + ( 8 - 1 ) d

a 8 = a 1 + 7 d = 18

a 15 = a 1 + ( 15 - 1 ) d

a 15 = a 1 + 14 d

a 9 = a 1 + ( 9 - 1 ) d

a 9 = a 1 + 8 d

a 15 - a 9 = ( a 1 + 14 d ) - ( a 1 + 8 d )

a 15 - a 9 = a 1 + 14 d - a 1 - 8 d

a 15 = a 1 - a1 + 14 d - 8 d

a 15 - a 9 = 6 d

a 15 - a 9 = 30

6 d = 30 Divide both sides by 6

d = 30 / 6

d = 5

a 8 = a 1 + 7 d = 18

a 1 + 7 * 5 = 18

a 1 + 35 = 18

a 1 = 18 - 35

a 1 = - 17

A.P

a n = - 17 + ( n - 1 ) * 5

- 17 , - 12 , - 7 , - 2 , 3 , 8 , 13 , 18 , 23 , 28 , 33 , 38 , 43 , 48 , 53

Answer 2

It is given that 4th term is 18 a4=18 a+3d=18
a=18-3d -(1)
a15-a9=a+14d-1-8d=30
6d=30
d=5
a=18-15=3
A.P is
3,8,13,18,23.....

Answer 3

Good enough to understand 💁👍👍

Answer 4

It is given that 4th term is 18
a4=18
a+3d =18
a=18-3d.......(1)
Also, a15 -a9 =a+14d-1-8d=30
6d= 30
d=5
From (1), we get
a= 18-15=3
Therefore the A.P is
3,8,13,18,23