1. **State the problem:** We are given the first three terms in ascending powers of $x$ in the expansion of $ (3 + px)^n $ as $243 + 810x + qx^2$. We need to find the values of $n$, $p$, and $q$. 2. **Recall the binomial expansion formula:** $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ The first three terms correspond to $k=0,1,2$: $$ T_0 = \binom{n}{0} 3^n (px)^0 = 3^n $$ $$ T_1 = \binom{n}{1} 3^{n-1} (px)^1 = n \cdot 3^{n-1} p x $$ $$ T_2 = \binom{n}{2} 3^{n-2} (px)^2 = \frac{n(n-1)}{2} 3^{n-2} p^2 x^2 $$ 3. **Match the given terms:** - Constant term: $3^n = 243$ - Coefficient of $x$: $n \cdot 3^{n-1} p = 810$ - Coefficient of $x^2$: $q = \frac{n(n-1)}{2} 3^{n-2} p^2$ 4. **Solve for $n$ from the constant term:** $$ 3^n = 243 $$ Since $243 = 3^5$, we get $$ n = 5 $$ 5. **Substitute $n=5$ into the coefficient of $x$ term:** $$ 5 \cdot 3^{4} p = 810 $$ Calculate $3^4 = 81$: $$ 5 \cdot 81 \cdot p = 810 $$ $$ 405 p = 810 $$ $$ p = \frac{810}{405} = 2 $$ 6. **Calculate $q$ using $n=5$ and $p=2$:** $$ q = \frac{5 \times 4}{2} \times 3^{3} \times 2^2 $$ Calculate each part: $$ \frac{5 \times 4}{2} = 10 $$ $$ 3^3 = 27 $$ $$ 2^2 = 4 $$ So, $$ q = 10 \times 27 \times 4 = 1080 $$ **Final answers:** $$ n = 5, \quad p = 2, \quad q = 1080 $$