In an given AP,a=7,d13=35,find d1and s13

Here, a = 7
a₁₃ = 35
a_n = a + (n - 1)d
{tex} \Rightarrow {/tex} a₁₃ = a + (13 - 1)d
{tex} \Rightarrow {/tex} a₁₃ = a + 12d
{tex} \Rightarrow {/tex} 35 = 7 + 12d
{tex} \Rightarrow {/tex} 12d = 35 - 7
{tex} \Rightarrow {/tex} 12d = 28
{tex} \Rightarrow d = \frac{{28}}{{12}}{/tex}
{tex} \Rightarrow d = \frac{7}{3}{/tex}
Again, we know that
{tex}{S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]{/tex}
{tex} \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + (13 - 1)d} \right]{/tex}
{tex} \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + 12d} \right]{/tex}
{tex}={S_{13}} = \frac{{13}}{2}\left[ {2(7) + 12\left( {\frac{7}{3}} \right)} \right]{/tex}
{tex} \Rightarrow {S_{13}} = \frac{{13}}{2}(14 + 28){/tex}
{tex} \Rightarrow {S_{13}} = \frac{{13}}{2}(42){/tex}
{tex} \Rightarrow {S_{13}} = (13)(21){/tex}
{tex} \Rightarrow {S_{13}} = 273{/tex}