# What is the value of a and b, given that 4+a, 8, 10b and a+2b are terms of AP

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4+ a , 8, 10b, a+2b are terms of an Arthematical progression:

Let (r) be the common difference:

Then we know that:

a1= 4+a

8 = (4+a) + r...........(1)

10b = (4+a) + 2r........(2)

a+ 2b= (4+a) + 3r.........(3)

Let us solve the system:

First we will rewrtie:

a + r = 4 . ==> a = 4-r...........(4)

10b = 4+ a + 2r

Substutue in with a:

==> 10b = 4 + (4-r) + 2r

==> 10b = 8 + r

==> r = 10b - 8..................(5)

a+2b = (4+a) + 3r

(4-r) + 2b = 4 + 4-r + 3r

2b = 4 + 3r .................(6)

2b = 4 + (10b-8)

2b = 4 + 10b - 8

==> -8b = -4

**==> b= 1/2**

==> r = 10b- 8

==> r= 5 - 8 = -3

**==> r= -3**

==> a = 4- r = 4 --3 = 7

**==> a = 7**

In an arithmetic progression, the difference between consecutive terms is the same. We use this property to build up two equations that can be solved to arrive at a and b.

10b - 8 = 8 - (4+a)

=> 10b – 8 = 8 – 4 – a

=> a = 12 – 10b … (1)

a+2b – 10b = 10b -8

=> a – 8b = 10b - 8

=> a = 18b - 8 … (2)

Equating the values of a from (1) and (2), we get

12 – 10b = 18b – 8

=> 28b = 20

=> b = 20/28 = 5/7

Now substitute b = 5/7 which we have determined earlier in (1)

a = 12 – 10*(5/7)

=> a = 12 – 50/7

=> a = 34/7

**Therefore the required values of a and b are 34/7 and 5/7 respectively.**

We suppose the terms , 4+a,8,10b, and a+2b are the consecutive terms of an arithmetic progressio.

So the difference of the duccessive terms must be equal to the common difference d.

8-(4+a) = 10b-8 .

4-a = 10b -8

-a-10b = = -8-4 = -12

a+10b = 12...(1).

Also

8-(4+a) = a+2b-10b.Or

4-a = a-8b

-a-a+8b = -4.

-2a+8b = -4

We divide by 2:

-a+4b = -2.....(2).

Eq(1)+eq(2) gives: 10b+4b = = 12-2 = 10.

14b = 10.

b= 10/14 = 5/7.

Put b= 1 in (2): -a +4*5/7 = -2.

-a = -2- 4*5/7 = -34/7

a = 34/7.

Therefore a= 34/7 and b = 5/7.