Question

the 3rd term of an arthamatic sequence in 34, and 6th term is 67 write the 1st 5 terms​

Answer 1

Answer:

12, 23, 34, 45, 56.

Step-by-step explanation:

3rd term = 34

So a + 2d = 34.

6th term = 67

So a + 5d = 67.

a + 2d = 34

a + 5d = 67

By eliminating a,

3d = 33

d = 11.

So a + 2(11) = 34.

a + 22 = 34

a = 12.

So the first 5 terms are 12, 23, 34, 45 and 56.

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