In an arithmetic sequence, t3 = 15 and s8 = 168, find the simplified expression of tn.?

a : the first term (T₁)
d : the common difference

T₃ = 15
a + (3 - 1)d = 15
a + 2d = 15 …… [1]

S₈ = 168
8 [2a + (8 - 1)d] / 2 = 168
2a + 7d = 42
a + 3.5d = 21 …… [2]

[2] - [1] :
1.5d = 6
d = 4

Substitute d = 4 into [1] :
a + 2(4) = 15
a = 7

Tn = a + (n - 1) d
Tn = 7 + (n - 1) × 4
Tn = 7 + 4n - 4
Tn = 4n + 3

t₁ = t₃ - (3-1)d = 15 - 2d
t₈ = t₃ + (8-3)d = 15 + 5d
t₁ + t₈ = 30 + 3d

S₈ = 8(t₁+t₈)/2
168 = 4(t₁+t₈)
t₁ + t₈ = 42

30 + 3d = 42
d = 4
t₁ = 15-2d = 7

nth term = t₁ + (n-1)d
 = 7 + (n-1)4
 = 7 + 4n - 4
 = 3 + 4n

General AP is a,a+d,a+2d,...,a+(n-1)d,... where a is 1st term, d is the common difference between consecutive terms {= t(n+1) - t(n), n = 1,2,3,...}, t(n), the nth term, = a+(n-1)d, n= 1,2,3,....& s(n), the sum of the first n terms, = (n/2)[t(1) + t(n)]. If you keep this definition in mind you will always be able to solve problems like this. Given t(3) = 15, we have a+(3-1)d = 15, ie., a+2d = 15..[1]. Given that s(8) = 168, we know n = 8.
Then s(8) = (8/2)[a+{a+(8-1)d}] = 4[2a+7d] = 168, ie., 2a+7d = 42..[2].
Now {[2]-2[1]}--> 3d = 12, ie., d = 4. Then [1]--> a = 15-8 = 7 & t(n) by
definition, = a+(n-1)d = 7 + (n-1)4 = 4n+3.

In an arithmetic sequence, t3 = 15 and s8 = 168.
Find the simplified expression of tn.
Assuming the cd to be 4, t1 = 7 and t8 = 35
7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 = 168
tn = 4n + 3

using the formula,

1. tn = a + (n-1) d
t3 = a + (3-1) d
15 = a +2d

2. sn= n/2 [2a + (n-1) d]
s8= 8/2 (2a + 7d)
168 = 8a + 28d

from 1.

a + 2d = 15
a = 15 - 2d

from 2.

8a + 28d = 168

use the 'a' from 1. then it will become like this.

8(15-2d) + 28 d = 168

120 - 16d + 28d = 168

120 + 12d = 168

12d = 168 -120
= 48
d= 48÷12
d = 4

a = 15 - 2(4)
a = 15-8
a = 7


Tn = a + (n - 1) d

Tn = 7 + (n -1) 4

Tn = 7+ 4n - 4


so the final answer is
Tn = 4n + 3


i don't know if i answer this right or wrong but i just feel like doing it anyway. i miss this subject even though the highest mark that i could get for this subject was 48 lol.