Arithmetc Progression containing odd terms Here is the question:

If for an AP of odd number of terms,the sum of all the terms is $\frac{15}{8}$ times the sum of the terms in odd places then find the number of terms in the AP.

my try:First of all i thought that in an odd AP there will be (2n+1) terms becuase (2n) terms will contain even number of terms where as there will be n+1 odd terms.
Then I equated the following equation
$\frac{2n+1}{2}[2a + 2nd]=\frac{15}{8}*\frac{n+1}{2}[2a+nd]$
But then I wondered there are three unknowns and one equation ,then how can i solve for $n$.So my question is that how can we solve for $n$ or $2n+1$ ?

First consider a constant progression, $t_n=1$. Then the ratio of the sums is $\dfrac{2n}{n+1}=\dfrac{15}8$ so that $\color{green}{n=15}$.
$$\frac{1+1+1+1+1+1+1+1+1+1+1+1+1+1+1}{1+1+1+1+1+1+1+1}=\frac{15}8.$$
This is compatible with a linear progression $t_n=n$, as
$$\frac{1+2+3+4+5+6+7+8+9+10+11+12+13+14+15}{1+3+5+7+9+11+13+15}=\frac{120}{64}=\frac{15}8.$$
By linearity, the property extends to any $t_n=a+bn$.
