Recently while solving few questions related to Progressions(specifically, A.P.), I realized one thing that in question like, "Find the sum of following series" and suppose the terms are up to $n^{th}$ places.
Then instead of going by the formula(s), I can simply add the first $3$ to $4$ given elements of the series and that would give me a rough idea of where the summation would belong. And since we are given Multiple Choice Question), I can solve it much more quickly.
So, is there any other trick that you may be aware of or formulated on your own related to solving any type of Progression? If yes, then please share it with us.
"Tricks" are always useful!
Anyway, here are some points for consideration.
P1. Positions in an AP or GP are relative
Think of $T_4-T_2=q-p$ as one "step".
$T_{10}$ is 3 steps away from $T_4$ (think $T_2, T_4, T_6, T_8, T_{10}$).
Hence $T_{10}=q+3(q-p)=4q-3p$.
In this way we avoid having to first work out $a$ and $d$ by solving simultaneous equations for $T_2, T_4$ and then applying the formula for $T_{10}$, which would be more laborious.
Similar, if the progression was a GP instead of an AP, then $$T_{10}=q\cdot \left(\frac qp\right)^3=\frac {q^4}{p^3}$$
P2. Visualize AP as a straight line graph
It might be helpful to visualize an AP as a straight line graph, with number of terms on the $x$-axis and value of each term on the $y$-axis.
The sum of the number of terms may be approximated by the area under the graph.
P3. A zero or negative term index may be a useful concept to provide symmetry
Taking into account P1 it may sometimes be useful to apply the concept of a zero or negative number-of-term index in order to provide symmetry to simplify the solution of a problem.
Let the three terms be $a-d, a, a+d$.
(NB: Here we consider $T_0, T_1, T_2$, instead of the usual $T_1, T_2, T_3$, WLOG).
Sum: $$(a-d)+a+(a+d)=3a=21\Rightarrow a=7$$ Product: $$(a-d)a(a+d)=a(a^2-d^2)=280\\ 7(49-d^2)=280\\ d=3$$ Hence the three terms are $4, 7, 10$.
This is much faster than solving for the terms $a, a+d, a+2d$ .