If aₙ = 3 - 4n, show that a₁, a₂, a₃,... form an AP. Also, find S₂₀
Solution:
Given, aₙ = 3 - 4n
We have to show that the given expression forms an AP and find the sum up to 20 terms.
Put n =1, a₁ = 3 - 4(1) = 3 - 4 = -1
Put n= 2, a₂ = 3 - 4(2) = 3 - 8 = -5
Put n= 3, a₃ = 3 - 4(3) = 3 - 12 = -9
Put n= 4, a₄ = 3 - 4(4) = 3 - 16 = -13
Put n= 20, a₂₀ = 3 - 4(20) = 3 - 80 = -77
So, the series is -1, -5, -9, -13,........, -77.
First term, a = -1
Last term, l = -77
Common difference, d = -5 - (-1) = -5 + 1 = -4
To check whether the series form an AP,
Common difference, d = -5 - (-1) = -9 - (-5) = -13 - (-9)
d = -5 + 1 = -9 + 5 = -13 + 9 = -4
d = -4
It is clear that the series is in AP.
If l is the last term of an AP, then the sum of the terms is given by
S = n/2[a+l]
So, S = 20/2[-1 + (-77)]
= 10[-5-77]
= 10(-78)
= -780
Therefore, the sum of the series up to 20 terms is -780.
✦ Try This: If aₙ = 4 - 3n, show that a₁, a₂, a₃,... form an AP. Also, find S₂₀
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 23
If aₙ = 3 - 4n, show that a₁, a₂, a₃,... form an AP. Also, find S₂₀
Summary:
An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same. If aₙ = 3 - 4n, the series -1, -5, -9,......., -77 form an AP. Also, S₂₀ is -780
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