# Arithmetic progression (given third term and difference between 5th and 7th term)

Given that the 3rd term of an arithmetic progression (AP) is 16 and the difference between the 5th and the 7th term is 12, write down the first 7 terms of the AP.

For an AP, the $n^{th}$ term is given by:

$$a_n=a+(n-1)d$$

where $a$ is the first term and $d$ is the common difference

The difference between the $5^{th}$ and the $7^{th}$ is $12$

$$12 / 2 = 6$$

$$a_3=a+2d=16$$

$$a_3=a+2(6)=16$$

$$a_1=(16-2(6))$$

$$a_1=(16-12) = 4$$

Is this the correct method to find the first term?

• But you wrote $a_1=(16-12) = 4$ so you already found the first term...? And since you found the difference to be $d=6$, it shouldn't be too hard to write down the following terms. – StackTD Jun 18 '18 at 13:31
• I have reworded my question. I was not sure if that was the correct methodology. Thanks. – Mike S Jun 18 '18 at 13:34
• Alright; this looks fine! – StackTD Jun 18 '18 at 13:36
• Is the 5th term greater or less than the 7th term? – hypergeometric Jun 18 '18 at 15:19

## 1 Answer

Assuming that the 7th term is greater than the 5th term, we have the following:

Note that $T_1, T_3, T_5, T_7$ are also in AP (AP2).

Given that $T_3=16$, and $T_7-T_5=12$ (i.e. common difference for AP2 is $12$), we have $$T_{1,3,5,7}=4,16,28,40$$.

Interpolating (since an AP is linear) we have

$$T_{1,2,3,4,5,6,7}=-2,4,10,16,22,28,34,40$$