# Q : 10    Show that  $\small a_1,a _2,...,a_n,...$ form an AP where an is defined as below :               (ii) $\small a_n=9-5n$                Also find the sum of the first $\small 15$ terms in each case.

G Gautam harsolia

It is given that
$\small a_n=9-5n$
We will check values of $a_n$ for different values of n
$a_1 = 9-5(1) =9-5= 4$
$a_2 = 9-5(2) =9-10= -1$
$a_3 = 9-5(3) =9-15= -6$
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 4 and common difference (d) equals to -5
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(4) +(15-1)(-5)\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ 8 -70\right \}$
$\Rightarrow S_{15}= \frac{15}{2}\left \{ -62\right \}$
$\Rightarrow S_{15}= 15 \times (-31)$
$\Rightarrow S_{15}= -465$
Therefore, the sum of 15 terms is -465

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