In an AP: a _ 12 = 37, d = 3, find a and S_ 12 . - Vidyadip

Question

In an AP: $a _{12} = 37, d = 3$, find a and $S_{12}.$

✓

Answer

Here, $a_{12} = 37$
$d = 3$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow a_{12} = a + (12 - 1)d$
$ \Rightarrow a_{12} = a + 11d$
$ \Rightarrow 37 = a + 33$
$ \Rightarrow a = 37 - 33 = 4$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{12}} = \frac{{12}}{2}\left[ {2a + (12 - 1)d} \right]$
$ \Rightarrow {S_{12}} = 6\left[ {2a + 11d} \right]$
$ \Rightarrow {S_{12}} = 6\left[ {2 \times 4 + 11 \times 3} \right]$
$ \Rightarrow {S_{12}} = 6\left[ {8 + 33} \right]$
$ \Rightarrow {S_{12}} = 6 \times 41$
$ \Rightarrow {S_{12}} = 246$

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