Question

Find the sum of arithmetic sequences 4+8+12+16+...+400

Answer 1

Answer:

20200

Step-by-step explanation:

a=1  ×  4=4

a=2  ×  4=8

a=3  ×  4=12

a = 4  × 4 = 16

400 = 4 + ( n - 1 ) 4]

400 = 4 + 4n - 4

4n = 400

400 / 4n

n = 100

100 / 2 [ 4 + 400 ]

50 [ 404 ]

50 × 404

= 20200

Hope this answer helps you :)

Have a great day

Mark brainliest

Answer 2

Answer:

Step-by-step explanation:

First term = a =4

Common difference = d = 2nd term - first term

d = 8 - 4 = 4

Last term = l = 400

First we have to fin the number of terms 'n'

l = a + (n-1)d

4 + (n-1)*4 = 400

4 + 4n- 4 = 400

      4n = 400     {Divide both sides by 4}

         n = 400/4

n = 100

$S_{n}=\frac{n}{2}(a + l)\\\\ = \frac{100}{2}*(4+400)\\\\=50 * 404\\\\= 20200$