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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?

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    $\begingroup$ 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. $\endgroup$ – StackTD Jun 18 '18 at 13:31
  • $\begingroup$ I have reworded my question. I was not sure if that was the correct methodology. Thanks. $\endgroup$ – Mike S Jun 18 '18 at 13:34
  • $\begingroup$ Alright; this looks fine! $\endgroup$ – StackTD Jun 18 '18 at 13:36
  • $\begingroup$ Is the 5th term greater or less than the 7th term? $\endgroup$ – hypergeometric Jun 18 '18 at 15:19
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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

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

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