# 2t+1, 12, 3t-2 are terms of A.P, find the value of t.

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For 3 terms of an AP the term in the middle is the average of the 1st and the 3rd terms.

As 2t + 1 , 12 and 3t - 2 are terms of an AP

=> (2t + 1 + 3t - 2)/2 = 12

=> (2t + 1 + 3t - 2) = 24

=> 5t - 1 = 24

=> 5t = 25

=> t = 25/5

=> t = 5

**The value of t is 5.**

Given the terms (2t+1) , 12, (3t-2) are terms of an arithmetical progression.

Let the common difference be r .

Then we know that:

2t+1 = 12 -r

3t-2 = 12 +r

We will add both equations.

==> 5t -1 = 24

==> 5t = 25

==> t= 5

To check the answer we will substitute into the terms.

==> 2t+1 = 2*5+1 = 11

==> 12

==> 3t-2 = 3*5-2 = 15-2 = 13

==> 11, 12, 13, are terms of A.P and the common difference is r= 1

**Then the value of t = 5**