# if 3, b, -1, c are terms of an A.P, then find b and c

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3, b , -1 , c are terms of an Arthematical progression.

Let r be the common difference:

Then we know that:

b = 3 + r.........(1)

-1 = 3 + 2r........(2)

c = 3 + 3r...........(3)

Let us solve equation (2) to determine r:

-1 = 3 + 2r

==> 2r= -4

==> r= -2

Then the common difference r = -2

Now let us substitue in (1):

b= 3+ r = 3+ -2 = 1

**==> b = 1**

Now we will substitute in (3):

c = 3 + 3r

c = 3+ 3*-2

==> c = 3-6 = -3

**==> c = -3**

**Then the series will be:**

**3, 1, -1, -3 is an A.P where r = -2 is the common difference**

If a1, a2 and a3 are the consecutive terms of an AP, then

2a2 = a1+a3.

Twice middle term a2 = a1 +a2.

Given that 3, b, -1, c are the terms (assumed consecutive terms) of an AP.

Therefore for the first 3 terms 2b = (3+1). So b= 4/2 = 2.

Having determined b = 2, for the second 3 terms: b, -1 , c or 2, -1, c, we have:

2*(-1) = b+c= 2+c.

-2 = 2+c.

So c = -2 -2 = -4.

Therefore c = -4,

So, b = -1 and c = -4.