If $(m+1)$ th ,$(n+1)$ th,$(r+1)$ th terms of a non constant AP are in GP and $m,n,r$ are in HP,then prove that the ratio of first term of the AP to its common difference is $\frac{-n}{2}$
Let$(m+1)$ th term of a non constant AP$=a+md$ where $a,d$ are the first term and the common difference of the AP
$(a+nd)^2=(a+md)(a+rd)$ and $\frac{2}{n}=\frac{1}{m}+\frac{1}{r}$
Now i am not able to find $\frac{a}{d}.$
Clearly, $(\frac ad +n)^2 = (\frac{a}{d} + m)(\frac ad +r)$, by dividing the first equation by $d^2$. Expanding and cancelling terms we get $\frac{2an}{d} + n^2 = \frac{a(m+r)}{d} + mr$.
Transposing terms, we have $\frac{a}{d}(2n-m-r) = mr-n^2$.
Consequently, $\frac ad = \frac{mr - n^2}{2n-m-r}$. Since we know the answer is $\frac{-n}{2}$, let us rewrite the above as $\frac{-n}{2} \times \frac{2mr - 2n^2}{n(m+r) - 2n^2}$, where we multiplied the numerator by $2$ and denominator by $-n$ to compensate.
Now, $n = \frac{2mr}{r+m}$, so using this we see that the second fraction is $1$ by simplifying the denominator as $n(m+r) = 2mr$. Hence we are done.