Given an arithmetic progression where the $$ p\text{-th term}=2p+4q-5r\\ q\text{-th term}=3p+3q-5r\\ r\text{-th term}=3p+4q-6r$$ determine the value of the $(p+q+r)$-th term.
Also, find $n$ where the $n$-th term equals zero.
__
NB: This question was designed as an extension of some similar questions here on MSE.
On
Note that adding $p,q,r$ back to the $p,q,r$-th terms respectively gives a constant $3p+4q-5r$.
Hence the $n$-th term of the AP is given by
$$U_n=3p+4q-5r-n$$
Putting $n=p+q+r$: $$\color{red}{U_{p+q+r}=2p+3q-6r}$$
From the definition of $U_n$ it is clear that for $U_n=0$, $$\color{red}{n=3p+4q-5r}$$
On
Let $a_i$ be the $i$-th term for $i\in\mathbb{N}$. Then observe that, $$a_{p+q+r}=a_1+(p+q+r-1)$$$$a_{p}=a_1+(p-1)$$$$a_{q}=a_1+(q-1)$$$$a_{r}=a_1+(r-1)$$Now observe that \begin{align}a_p+a_q+a_r&=3a_1+(p+q+r-3)\\&=3a_1+3(p+q+r-3)-2(p+q+r-3)\\&=3a_1+3(p+q+r-1)-6-2p-2q-2r+6\\&=3a_{p+q+r}-2(p+q+r)\end{align}Consequently, $$\boxed{a_{p+q+r}=\dfrac{a_p+a_q+a_r+2(p+q+r)}{3}}$$
You can, use the above argument to prove a more general problem. Namely,
Given an arithmetic progression of $n$-terms (the $i$-th term being denoted as ${\bf{T}}(i)$ with $i\in\mathbb{N}$ if $p_1,p_2,\ldots,p_m$-th terms are known (where $m,p_1,p_2,\ldots,p_m\in\mathbb{N}$ and $m\le n$) then what is ${\bf{T}}\left(\displaystyle\sum_{i=1}^n p_i\right)$?
Hint: Prove that, $$\boxed{\displaystyle\sum_{i=1}^m {\bf{T}}(p_i)=m{\bf{T}}\left(\displaystyle\sum_{i=1}^mp_i\right)-(m-1)\left(\displaystyle\sum_{i=1}^m p_i\right)}$$
We have $$a_p = a_1 + (p-1)d = 2p+4q-5r...(1)$$ $$a_q = a_1 + (q-1)d = 3p+3q-5r...(2)$$ $$a_r =a_1 + (r-1)d = 3p+4q-6r...(3)$$ Now, subtracting $(1)$ and $(2)$ gives us $d=-1$ and thus $a_1 = 3p+4q-5r-1$.
Then, $$a_{p+q+r} = a_1 + (p+q+r-1)d = 3p+4q-5r-1 -p-q-r+1 = 2p+3q-6r$$ If $a_n = 0$, then $$a_1 + (n-1)d = 0 \Rightarrow n = a_1 + 1=(3p+4q-5r)^{\text{th}} \text{ term }$$