If an = 3 – 4n, then show that a1, a2, a3, … from an AP. Also, find S20.


Given that, nth term of the series is an = 3 - 4n


For a1,


Put n = 1 so a1 = 3 - 4(1) = - 1


For a2,


Put n = 2, so a1 = 3 - 4(2) = - 5


For a1,


Put n = 3 so a1 = 3 - 4(3) = - 9


For a1,


Put n = 4 so a1 = 3 - 4(4) = - 13


So AP is - 1, - 5, - 9, - 13, …


a2 - a1 = - 5 - (- 1) = - 4


a3 - a2 = - 9 - (- 5) = - 4


a4 - a3 = - 13 - (- 9) = - 4


Since, the each successive term of the series has the same difference. So, it forms an AP with common difference, d = - 4


We know that, sum of n terms of an AP is



Where a = first term


d = common difference


and n = no of terms



= 10[ - 2 - 76]


= - 780


So Sum of first 20 terms of this AP is - 780.


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