The fourth term of an Arithmetic sequence is 18 and the sixth term is 28. Give the first 3 terms?

Let a be the first term of the arithmetic sequence, and d be the common difference.

T(4): a + 3d = 18 ...... [1]
T(6): a + 5d = 28 ...... [2]

[2] - [1] :
2d = 10
d = 5

Substitute d = 5 itno [1] :
a + 3(5) = 18
a = 3

The 1st term, T(1) = 3
The 2nd term, T(2) = 3 + 5 = 8
The 3rd term, T(3) = 8 + 5 = 13

The first three terms are 3, 8 and 13.

Let a be the first term and d be the common difference
a+3d = 18
a+5d = 28

subtract
-2d = -10
d = 5
a+3d = 18
a+3(5) = 18
a+15 = 18
a = 3

a= 3 and d = 5
The first three terms are a, a+d, a+2d
3, 8, 13

The gap of 10 between 18 and 28 is twice the common difference which is 5. Subtract 5s from 18 and work backwards to the first term:
13
8
3

Well,

a4 = a + 3d = 18 (1)
a6 = a + 5d = 28 (2)
therefore (2) - (1) gives
2d = 28 - 18 = 10 ==> d = 5
and
a4 = a + 3d = 18 and d=5 ==> a = 3
conclusion
a1 = 3
a2 = 8
a3 = 13

hope it' ll help !!

"Arithmetic" sequence = the same value is ADDED at each step.

Call this value "x" for now.

4th term is 18
5th term is 18 + x
6th term is 18 + x + x = 18 + 2x + 28

obviously, the value added at every step is 5

Now, simply go backwards,knowing that the step value is 5

4th term = 18
3rd term = 13
2nd term = 8
1st term = 3

(0th term = -2, often used by mathematicians to reconstruct a formula)

If "n" represents the term number, then the "n"th term is

-2 + 5n

What is the sixth term (n = 6)
-2 + 5(6) = -2 + 30 = 28

What is the 100th term?
-2 + 5(100) = -2 + 500 = 498