# 2, a, b, 17 are terms in A.P. find a and b.2, a, b, 17 are terms in A.P. find a and b.

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2, a, b , 17 are terms is an arithmetical progression.

The:

a1= 2

a2= a1+ r = 2 + r = a

a3= a1+ 2r = 2 + 2r = b

a4= a1+ 3r = 2+ 3r = 17

==> 3r = 15

==> r = 5

==> a= 2+r = 2+5 = 7

==> b = 2+ 2r = 2+2*5 = 12

==> 2, 7, 12, 17 are terms in an A.P where the constant difference r is 5.

Because the terms of the sequence are the terms of an arithmetic sequence, we'll use the arithmetic mean theorem to calculate a and b.

For instance:

a = (2+b)/2 => 2a = 2+b => 2a - b = 2 (1)

b = (a+17)/2 => 2b = a+17 => a = 2b - 17 (2)

We'll substitute (2) in (1) and we'll get:

2*(2b-17) - b = 2

We'll remove the brackets:

4b - 34 - b - 2 = 0

3b = 36

We'll divide by 3:

**b = 12**

We'll substitute b in (1):

2a - 12 = 2

2a = 14

We'll divide by 2:

**a = 7**

We'll verify if th values of a and b are proper calculated, so we'll find out the common difference, d:

a-2 = d, where a = 7

7-2 = 5 => d = 5

b-a = 5

12 - 7 = 5

5 = 5

**So, a=7 and b=12 are the terms of the given arithmetic sequence.**

2,a,b,17 are in AP.

To find a and b.

Solution:

a1 = 2,

a2 =a

a3 = b

a4= 17.

Since this is an AP, the rth term , ar = a1+(r-1)d, where d is the constant common increment between any two succesive terms.

a4 = a1+3d = 17

a1 = a1 = 2.

a4 - a1 = 3d = 17-2 =15

3d = 15

3d/3 = 15/3 =5

d = 5.

So a2 = a = a1+5 = 2+5 =7

a3 =b = a1+2d = 2+2*5 = 12

Let:

The common difference between consecutive terms in the given A.P. = d

Then the various terms of the A.P> can be represented as:

First term = 2

Second term = a = 2 + d ... (1)

Third term = b = 2 + 2d .... (2)

Fourth term = 17 = 2 + 3d ... (3)

Simplifying equation (3) we get:

3d = 17 - 2 = 15

Therefore:

d = 15/3 = 5

Substituting this value of d in equation (1):

a = 2 + d

= 2 + 5 = 7

And substituting value of d in equation (2):

b = 2 + 2d

= 2 + 2*5

= 2 + 10 = 12

Answer:

a = 7

b = 12

2,a,b,17 are consecutive terms in an Arithmetic Progression.

Therefore 2+b=2*a ......(1)

and a+17=2*b ......(2)

Let's solve (1) and (2)

Multiply (1) by 2 and add (2) we get

4+2*b+a+17=4*a+2*b

=>3*a=21

=>a=7

Substitute a=7 in (2) 7+17=2*b or b=24/2=12

Therefore a=7 and b=12