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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
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.
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