Question
In an AP: $a = 7, a_{13} = 35$, find d and $S_{13}.$
✓
Answer
Here, $a = 7$
$a_{13} = 35$
$a_n= a + (n - 1)d$
$ \Rightarrow a_{13} = a + (13 - 1)d$
$ \Rightarrow a_{13} = a + 12d$
$ \Rightarrow 35 = 7 + 12d$
$ \Rightarrow 12d = 35 - 7$
$ \Rightarrow 12d = 28$
$ \Rightarrow d = \frac{{28}}{{12}}$
$ \Rightarrow d = \frac{7}{3}$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + (13 - 1)d} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + 12d} \right]$
$={S_{13}} = \frac{{13}}{2}\left[ {2(7) + 12\left( {\frac{7}{3}} \right)} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(14 + 28)$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(42)$
$ \Rightarrow {S_{13}} = (13)(21)$
$ \Rightarrow {S_{13}} = 273$
$a_{13} = 35$
$a_n= a + (n - 1)d$
$ \Rightarrow a_{13} = a + (13 - 1)d$
$ \Rightarrow a_{13} = a + 12d$
$ \Rightarrow 35 = 7 + 12d$
$ \Rightarrow 12d = 35 - 7$
$ \Rightarrow 12d = 28$
$ \Rightarrow d = \frac{{28}}{{12}}$
$ \Rightarrow d = \frac{7}{3}$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + (13 - 1)d} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + 12d} \right]$
$={S_{13}} = \frac{{13}}{2}\left[ {2(7) + 12\left( {\frac{7}{3}} \right)} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(14 + 28)$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(42)$
$ \Rightarrow {S_{13}} = (13)(21)$
$ \Rightarrow {S_{13}} = 273$
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