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

Question

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

in progress 0
Dulcie 4 years 2021-08-09T12:20:47+00:00 2 Answers 37 views 0

Answers ( )

  1. linhdan
    0
    2021-08-09T12:22:11+00:00
    Reply

    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

  2. Maris
    0
    2021-08-09T12:22:37+00:00
    Reply

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )