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.

Answers (1)

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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