I can't seem to figure this out! What is the value of the 129th term in a sequence in which the 17th term has a value of 28 and the 33rd term has a value of 78?

Step-by-step explanation:

The nth term of an Arithmetic sequence is expressed as;

Tn = a + (n-1)d

a is the first term

n is the numebr of terms

d is the common difference

If the 17th term has a value of 28, then

T17 = a + 16d = 28 1

If the 33rd term has a value of 78 then;

T33 = a+32d = 78 2

Solve simultaneously;

a + 16d = 28 1

a+32d = 78 2

Subtract

16d - 32d = 28-78

16d = 50

d = 50/16

d = 25/8

From 1;

a + 16d = 28

a + 16(25/8) = 28

a + 50 = 28

a = 28 - 50

a = -22

Get the 129th term

T129 = a + 128d

T129 = -22 + 128(25/8)

T129 = -22 +25(16)

T129 = -22+400

T129 = 378

Hence the 129th term is 378

with what

Step-by-step explanation:

there's no question