Answer:
Given, sum of first 7 terms of an A.P \[({{S}_{7}})=49\] and sum of first 17 terms of an A.P \[({{S}_{17}})=289\] i.e., \[{{S}_{7}}=\frac{7}{2}[2a+(7-1)d]=49\] \[2a+6d=14\] ?(i) And. \[{{S}_{17}}=\frac{17}{2}[2a+(17-1)d]=289\] \[2a+16d=34\] ?(ii) Solving equations (i) (ii), we get \[_{\begin{smallmatrix} \,\,2a\,+\,6d\,\,=\,14 \\ -\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,- \\ \overline{\,\,\,\,\,\,\,\,\,\,\,10\,d\,=\,\,20} \end{smallmatrix}}^{2a\,+\,16\,d\,=\,34}\] \[d=2\] Putting \[d=2\] in eq. (i), we get \[a=1\] Hence, sum of first n term of A.P, \[{{S}_{n}}=\frac{n}{2}[2(1)+(n-1)2]\] \[\Rightarrow {{S}_{n}}=\frac{n}{2}[2+(n-1)2]\] \[\Rightarrow {{S}_{n}}={{n}^{2}}\]
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