3, -2, a, b, -17 are terms of an A.P find a and b

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3, -2, a, b, -17 are terms of an A.P

Let us calculate the common difference (r).

r= -2 - 3 = -5

==> r= -5

Now we know that:

a3 =a = a1 + 2*r

= 3 + 2r

= 3 + 2*-5

= 3 - 10 = -7

**==> a= -7**

also:

a4 = b = a1+ 3*r

= 3+ 3*-5

= 3 -15 = -12

**==> b = -12**

Then ,

**3, -2, -7, -12, -17 is an A.P with r = -5**

We are given that 3, -2, a, b, -17 are the terms of an A.P.

Now for terms of an AP if we have three consecutive terms a1 , a2 and a3 we have a1+ a3 = 2* a2

Using the terms of the series here we have

3+ a = -2*2 = -4

-17 + a = 2b

From 3 + a = -4, we get a = -7

substitute a = -7 in -17 + a = 2b

=> -17 -7 =2b

=> b = -12

**Therefore a = -7 and b - 12**

Since 3,-2,a, b and -17 are all in AP, the common difference between the succesive terms should be same:

-2-3 = a - -2 = b-a = -17-b.

-5 = a- -2 = a+2

-5-2 = a.

a = -7

a- -2 = b-a

-7+2 = b- -7 = b+7

-7-7 +2 = b

b = -12.

Therefore a = -7 and b = -12. and the common difference = -5

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