If the sum of 7 terms of an AP is 49 and that of 17 terms is 289,

Solution:

$\mathrm{S}_{7}=49$

$\Rightarrow \frac{7}{2}\{2 \mathrm{a}+6 \mathrm{~d}\}=49 \Rightarrow \mathrm{a}+3 \mathrm{~d}=7 \ldots(1)$

$\mathrm{S}_{17}=289$

$\Rightarrow \frac{17}{2}\{2 a+16 d\}=289 \Rightarrow a+8 d=17 \ldots(2)$

Subtracting (1) from (2), we get

$5 \mathrm{~d}=17-7=10$

$\Rightarrow d=2$

From (1),

a + 3 × 2 = 7

$\Rightarrow \mathrm{a}=1$

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}\}$

$=\frac{n}{2}\{2 \times 1+(n-1) \times 2\}$

$=\frac{n}{2}\{2 n\}=n^{2}$

Hence, $S_{n}=n^{2}$