Question about arithmetic progression. stuck on one of the answers

Hint:

Remember that an arithmetic progression is a series where two consecutive terms have the same common difference.

Also, note that $a_n,a_{n+1}$ are two consecutive terms and $a_{n-1},a_n$ are two consecutive terms.

So, by definition, $a_n-a_{n-1}$ and $a_{n+1}-a_n$ should be constant common difference.

However, for the second question, the common difference is dependent on $n$ and thus not constant. So the second one is not an AP.

Hoewvwer an alternate approach to solve the second problem exists.

Using the fact $a_n-a_{n-1}=2n-1$ we get,

$$a_2-a_1=3$$ $$a_3-a_2=5$$ $$a_4-a_3=7$$ $$.$$ $$.$$ $$a_n-a_{n-1}=2n-1$$

Adding the two columns we have $$a_n-a_1=3+5+7+...+(2n-1)$$ $$\implies a_n=1+3+5+7+...+(2n-1)\text{ [since a_1=1]}$$

So now the right side is an AP.

Add it and voila!

One way is to note, that that because n is variable, differences are in arithmetic progression in this case. This means, we can sum up all the differences as in an arithmetic progression. $n-1$ terms gives us ${(n-1)(3+(2n-1))\over 2}$ we can then add this to $a_1$ getting $1+(n-1)(n+1)= 1+n^2-1=n^2$