# Q : 10   Show that  $\small a_1,a_2,...,a_n,...$  form an AP where an is defined as below :               (i)   $\small a_n=3+4n$               Also find the sum of the first $\small 15$ terms .

It is given that
$\small a_n=3+4n$
We will check values of $a_n$ for different values of n
$a_1 = 3+4(1) =3+4= 7$
$a_2 = 3+4(2) =3+8= 11$
$a_3 = 3+4(3) =3+12= 15$
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to and common difference (d) equals to 4
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(7) +(15-1)4\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 14 +56\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 70\right \}$
$\Rightarrow S_{15}= 15 \times 35$
$\Rightarrow S_{15}= 525$
Therefore, the sum of 15 terms  is  525

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