# The third and fifth terms of an a.p are x+y respectively. find the 12th term.

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Let a1, a2, a3, ...., an is an A.P

a1 = a1

a2= a1 + r

a3 = a2 + r = x.........(1)

a4 = a3 + r = y........(2)

Now we thet a4 -a3 = r

==> **r = y-x **

a12 = a1 + 11r = a2 + 10r = a3 + 9r

==> a12 = x + 9*(y-x)

==> a12 = x + 9y - 9x

==> **a12 = 9y - 8x **

To calculate the terms of an a.p., we'll use the formula of a general term of an arithmetic sequence.

an = a1 + (n-1)r, where:

a1 is the first term

n is the number of terms

r is the common difference.

Now, we'll write a3, according to this formula:

a3 = a1 + (3-1)r, where a3 = x (from enunciation)

x = a1 + 2r (1)

a5 = a1 + (5-1)r, where a5 = y (also, from enunciation)

y = a1 + 4r (2)

We'll subtract (2) from (1):

a1 + 2r - a1 - 4r = x-y

We'll eliminate like terms:

-2r = x-y

We'll divide by -2:

**r = (y-x)/2**

So, the common difference is (y-x)/2.

Now, we can find a1.

a3 = a1 + 2r

x = a1 + 2(y-x)/2

x = a1 + y - x

a1 = x-y+x

**a1 = 2x - y**

Now, we have all the necessary elements to calculate a12:

a12 = a1 + 11r

We'll substitute a1 and r:

a12 = 2x-y + 11(y-x)/2

a12 = (4x - 2y + 11y - 11x)/2

We'll combine like terms:

**a12 = (-7x + 9y)/2**

The 3d and 5 terms of and arithmetic progression (AP) are x and y respectively.

To find the 12 th term.

Solution:

In an AP , the rth term is given by:

ar = a1+(r-1)d, where d is the common diffrence between the consecutive terms.

Given a3 = a1+(3-1)d = x. Or

a1+2d = x.......(1)

a5 = a1+(5-1)d = y. Or

a1+4d = y.........(2)

(2) -(1) gives:

(a1+4d)-(a1+2d) = y-x

a1+4d-a1-2d = y-x.

4d-2d = y-x

2d = y-x

d = (y-x)/2 is the common difference.

Therefore a1 = a3- (3-1)d =

a1 = a3-2d

a1 = x - 2(y-x)/2

a1 = x- (y-x) = 2x-y.

So the 12 th term a12 = a1+(12-1)d

a12 = a1+11d

a12 = a1+11(y-x)/2

a12 = 2x-y +11(y-x)/2

a12 = 2x+11y/2-11x/2

a12 = (2-11/2)x/2 + 11y/2

a12 = -7x/2+ 11y/2

a12 = (11y-7x)/2 is the 12 term.