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