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$\\Solution:\\ \text{Let a be the first term and d be the common difference of given AP}\\ \text{Given n th term of an AP}=3-2n\\ So,\\ \text{1st term of an AP (a)}=3-2\times1=3-2=1\\ \text{2nd term of an AP=}=3-2\times 2=3-4=-1\\ Common\ difference\ (d)=2nd\ term-1st\ term=-1-1=-2\\\\\\ We\ know\ \text{Sum of first n term of an AP}= \frac{n}{2}(2a+(n-1)d)\\\\ \text{Sum of first 15 term of an AP }(S_{15})= \frac{15}{2}(2\times1+(15-1)(-2))\\\\\\ \Rightarrow S_{15}=\frac{15}{2}(2+14\times-2)\\\\\\ \Rightarrow S_{15}=\frac{15}{2}(2-28)=\frac{15}{2}\times(-26)\\\\\\ \therefore S_{15}=-195$

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