# What are the first term and the common differenceof an A.P. if the 2nd term is 5 and sixth term is 17?

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Given that the second term is 5 and the 6th term is 17 in an A.P.

Let the common difference be r and the first term is a1.

Then we know that:

a2= a1+ r

==> 5 = a1+ r...............(1)

Also we know that:

a6 = a1+ 5r

==> 17 = a1+ 5r ..............(2)

Now we will solve the system .

We will subtract (1) from (2).

==> 12 = 4r

==> r= 3

Now we will substitute into (1) to find a1.

==> 5= a1+ r

==> 5= a1+ 3 ==> a1= 5-3 = 2

**Then the first term is a1=2 and the common difference is r=3**

We'll recall the identity that expresses the general term of an arithmetic progression:

an=a1 + (n-1)d, where a1 is the first term and d is the common difference.

a2=a1 + (2-1)d

a6=a1 + (6-1)d

We'll substitute a2 and a6 by the values given in enunciation:

5 = a1 + d

17 = a1 + 5d

We'll subtract the 1st relation from the 2nd and we'll get:

17 - 5 = a1 + 5d - a1 - d

We'll eliminate and combine like terms:

12 = 4d

d=3

We'll substitute d in the first relation:

5 = a1 + d

5 = a1 + 3

a1= 5 - 3

a1 = 2

**We've found out the values of the first term and the common difference of the given a.p. as being: a1 = 2 and d = 3.**