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(C) Given $a_{n}=9-5n$.Calculating the first few terms:$a_{1}=9-5(1)=4$$a_{2}=9-5(2)=-1$$a_{3}=9-5(3)=-6$Checking the common difference:$a_{2}-a_{1}=-

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$-465$

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(C) Given $a_{n}=9-5n$.Calculating the first few terms:$a_{1}=9-5(1)=4$$a_{2}=9-5(2)=-1$$a_{3}=9-5(3)=-6$Checking the common difference:$a_{2}-a_{1}=-1-4=-5$$a_{3}-a_{2}=-6-(-1)=-5$Since the difference $a_{k+1}-a_{k}$ is constant $(-5)$,the sequence forms an $AP$ with first term $a=4$ and common difference $d=-5$.The sum of the first $n$ terms is given by $S_{n}=\frac{n}{2}[2a+(n-1)d]$.For $n=15$:$S_{15}=\frac{15}{2}[2(4)+(15-1)(-5)]$$S_{15}=\frac{15}{2}[8+14(-5)]$$S_{15}=\frac{15}{2}[8-70]$$S_{15}=\frac{15}{2}(-62)$$S_{15}=15(-31)=-465$.

Show that $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ form an $AP$ where $a_{n}$ is defined as $a_{n}=9-5n$. Find the sum of the first $15$ terms.

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