Q.Compute the sum of first $n$ terms of the following series: i. $8+88+888+8888+\cdots$ ii. $6+66+666+6666+\cdots$

A finite series is the sum of the terms of a finite sequence: if $(a_n)$ is a sequence, $a_1+a_2+\cdots+a_n=\sum_{k=1}^n a_k$ is the corresponding finite series (sometimes indexed to start from $a_0$).

Sum of an AP. For $T_n=a+(n-1)d$,

Sum of a GP. For $T_n=ar^{n-1}$ ($r\ne1$),

if $r=1$ the GP is the constant sequence $a,a,\ldots$ and $S_n=na$. Equivalently $1+r+r^2+\cdots+r^{n-1}=\dfrac{1-r^n}{1-r}$ for $r\ne1$.

Sum of an AGP. For $T_n=(a+(n-1)d)r^{n-1}$ ($r\ne1$),

This is derived the same way as the GP-sum formula: write $S_n$ and $rS_n$, subtract, and simplify the resulting mixed AP/GP remainder.

Telescopic summation. Many series that are not AP, GP or AGP can still be summed exactly if the $k^{th}$ term can be rewritten as a difference $t_k=f(k)-f(k+1)$ (often via rationalising a surd denominator, or a partial-fraction split). Then

since every interior term cancels against its neighbour and only the very first and very last survive — the "telescope" collapses. Two classic patterns: rationalising $\dfrac1{\sqrt k+\sqrt{k+1}}$ into $\sqrt{k+1}-\sqrt k$, and splitting $\dfrac1{k(k+1)}$ into $\dfrac1k-\dfrac1{k+1}$.

Word-problem recognition. A quantity that grows by adding a fixed amount each step (repayments increasing by a fixed rupee amount, balls spaced at equal intervals) is an AP — total distance/amount = $S_n$. A quantity that grows by a fixed multiple each step (bacteria doubling, compound interest, a virus doubling daily) is a GP — the amount AT a given step is $T_n=ar^{n-1}$ (a sequence question), while the amount ACCUMULATED over several steps is $S_n$ (a series question); reading the word problem carefully tells you which is being asked.