Question

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

Answer 1

Answer:

3+5+7+..........+401

d=a2-a1

d= 5-3

d = 2

an = a+ (n-1)d

401 = 3+(n-1)2

401=3+2n-2

401=2n+1

2n=401-1

2n=400

n=400/2

n=200

Sn= n/2 (a+an)

S200 = 200/2 (3+401)

=100(404)

Sum = 40400

hope it may help you........

Answer 2

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.