arithmetic

In an AP the first term is 2 and the sum of the first five terms is one fourth of the sum of the next five terms. Show that the 20th term is -112.

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  1. nth term is a+(n-1)d
    sum of 1st n terms = n/2(T1+Tn)
    a = 2

    sum of 1st 5 terms = 5/2 (2 + 2+4d) = 10+10d

    6th term = 2+5d
    sum of terms 6-10 = 5/2 (2+5d + 2+9d) = 10+35d

    10+10d = 1/4 (10+35d)
    40+40d = 10 + 35d
    5d = -30
    d = -6

    sequence is 2 -4 -10 -16 -22 -28 -34 -40 -46 -52

    sum of T1..T5 = -50
    sum of T6..T10 = -200

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    2. 👎
  2. nth term is a+(n-1)d
    sum of 1st n terms = n/2(T1+Tn)
    a = 2

    sum of 1st 5 terms = 5/2 (2 + 2+4d) = 10+10d

    6th term = 2+5d
    sum of terms 6-10 = 5/2 (2+5d + 2+9d) = 10+35d

    10+10d = 1/4 (10+35d)
    40+40d = 10 + 35d
    5d = -30
    d = -6

    sequence is 2 -4 -10 -16 -22 -28 -34 -40 -46 -52

    sum of T1..T5 = -50
    sum of T6..T10 = -200
    (copied From above)

    20th Term = a +(n-1)d

    = 2+19 X -6
    = 2 - 114
    = -112

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  3. 2ab/a² b²

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