# we have -4,t, 3t +1 are terms of an A.P. what is t?

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Given that -4, t, 3t+1 are terms in an arithmetical progression.

Let us assume that the common difference between terms is (r).

Then we know that:

a1= -4

a2= a1+ r = -4 + r = t

==> r - 4 = t

==> t= r-4 ....................(1)

Also, we know that:

a3= a1+ 2r

==> -4 + 2r = 3t+1

==> 2r = 3t + 5 .............(2)

Now we will substitute (1) into (2).

==> 2r = 3t + 5

==> 2r = 3(r-4) + 5

==> 2r = 3r - 12 + 5

==> -r = -7

==> r= 7

==> t= r-4 = 7-4 = 3

==>** t= 3.**

Let us verify our answer.

**==> -4 , t , 3t+1 = -4, 3, 10 are terms of an A.P where r = 7.**

We consider the 3 terms -4, t and 3t+1 to be consecutive terms of an AP.

Then the successive terms in the A.P. have the same common difference.

So t-(-4) = 3t+1-t.

=> t+4 = 2t+1.

=> 4 -1 = 2t-t.

=> 3 = t.

Therefore t= 3 .