3+5+7+..........................+401 Please find the sum.......... .

1.	3+5+7+..........................+401 Please find the sum.......... .
Answer» Given: A SUM of series is given which is in AP. The series is 3+5+7+9+....401. To Find: The sum of the series. Answer: Here the given series is , → 3+5+7+9+..............+401. And from the AP , First term = 3 Common Difference = (5-3)=2. Now we must find which term is 401 in ORDER to find the sum of series. $\large{\underline{\boxed{\red{\bf{\leadsto T_{n}=a+(n-1)d}}}}}$ $\sf{\implies T_{n}=3+(n-1)2}$ $\sf{\implies 401=3+(n-1)2}$ $\sf{\implies 401-3=(n-1)2}$ $\sf{\implies 398=(n-1)2}$ $\sf{\implies (n-1)=\dfrac{398}{2}}$ $\sf{\implies (n-1)=199}$ $\sf{\implies n =199+1}$ ${\underline{\boxed{\red{\bf{\longmapsto n=200}}}}}$ Hence the total number of terms is 200. $\rule{200}5$ Now sum of n terms of an AP is given by, $\large{\underline{\boxed{\red{\bf{\leadsto S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}}}$ $\sf{\implies S_{n}=\dfrac{200}{2}[2\times3+(200-1)2]}$ $\sf{\implies S_{n}=100[6+199\times2}$ $\sf{\implies S_{n}=100[6+398]}$ $\sf{\implies S_{n}=100\times 404}$ ${\underline{\boxed{\red{\bf{\longmapsto S_{n}=40400}}}}}$ Hence the required ANSWER is 40400.

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Given: