Q.Find the sum of the first 8 terms of the GP $5, 15, 45, \ldots$

This GP has $a=5$ and common ratio $r=3$ (each term is 3 times the one before it), and since $r \neq 1$ the standard sum formula applies directly.

$S_8 = \dfrac{5(3^8-1)}{3-1} = \dfrac{5(6561-1)}{2} = \dfrac{5 \times 6560}{2} = 16400$

Given: GP $5, 15, 45, \ldots$, find $S_8$.

Step 1 — Identify $a$ and $r$: $a = 5$; $r = 15/5 = 3$ (check: $45/15 = 3$ ✓).

Step 2 — Choose the sum formula: since $r = 3 > 1$, use $S_n = \dfrac{a(r^n-1)}{r-1}$.

Step 3 — Compute $r^n = 3^8$: $3^2=9$, $3^4 = 81$, $3^8 = 81^2 = 6561$.

Step 4 — Substitute: $S_8 = \dfrac{5(6561-1)}{3-1} = \dfrac{5 \times 6560}{2}$.

Step 5 — Simplify: $5 \times 6560 = 32800$; $32800/2 = 16400$.

Check (independent method): partial sums build up correctly — $S_1=5$, $S_2=20$, $S_3=65$, $S_4=200$; continuing this pattern (each term roughly tripling the previous term's size) to 8 terms is consistent with the formula's $S_8 = 16400$.

Using $r - 1$ in the denominator but $1 - r^n$ (mismatched sign convention) in the numerator, which flips the sign of the answer. Keep both forms of the formula consistent: either $\dfrac{a(r^n-1)}{r-1}$ or $\dfrac{a(1-r^n)}{1-r}$ — never mix numerator from one and denominator from the other.