If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Step-by-Step Solution
Key Concept: Use the formula for the sum of first $n$ terms of an arithmetic progression, $S_n = \frac{n}{2}[2a+(n-1)d]$, to set up two equations for the given sums, solve for the first term $a$ and common difference $d$, and then substitute back to obtain a general expression for $S_n$.
1. Write the sum formula for the given numbers of terms.
\[
S_7 = \frac{7}{2}[2a+(7-1)d] = 49 \quad\text{and}\quad S_{17}=\frac{17}{2}[2a+(17-1)d] = 289.
\]
2. Simplify each equation.
- For $S_7$: \[\frac{7}{2}(2a+6d)=49 \Rightarrow 7(2a+6d)=98 \Rightarrow 2a+6d=14 \Rightarrow a+3d=7 \quad\text{(Equation 1)}\]
- For $S_{17}$: \[\frac{17}{2}(2a+16d)=289 \Rightarrow 17(2a+16d)=578 \Rightarrow 2a+16d=34 \Rightarrow a+8d=17 \quad\text{(Equation 2)}\]
3. Subtract Equation 1 from Equation 2 to find $d$.
\[ (a+8d)-(a+3d)=5d = 17-7 = 10 \]
Hence, \[ d = \frac{10}{5}=2. \]
4. Find $a$ using Equation 1.
\[ a+3d = 7 \Rightarrow a+3(2)=7 \Rightarrow a+6=7 \Rightarrow a = 1. \]
5. Write the general sum $S_n$ using the found $a$ and $d$.
\[ S_n = \frac{n}{2}[2a+(n-1)d] = \frac{n}{2}[2(1)+(n-1)\cdot2]. \]
Simplify:
\[ S_n = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}[2n] = n^2. \]
6. Result: The sum of the first $n$ terms of the AP is \[ \boxed{S_n = n^2}. \]
Correct Answer: S_n = n^2