The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n
Solution:
Given,
The sum of first n terms of an AP is equal to the sum of first 2n terms of another AP.
We have to find the value of n.
given, the first AP series have
First term, a = 8
Common difference, d = 20
Also, the second AP series have
First term, a = -30
Common difference, d = 8
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n-1)d]
For the first series, Sₙ = n/2[2(8) + (n - 1)(20)]
= n/2[16 + 20n - 20]
= n/2[20n - 4]
Taking out common term,
= (n/2)2[10n - 2]
Cancelling out common term,
Sₙ = n[10n - 2]
For the second series, S₂ₙ = (2n/2)[2(-30) + (2n - 1)(8)]
= n[-60 + 16n - 8]
S₂ₙ = n[16n - 68]
Given, Sₙ = S₂ₙ
n[10n - 2] = n[16n - 68]
10n - 2 = 16n - 68
10n - 16n = -68 + 2
-6n = -66
n = 66/6
n = 11
Therefore, the value of n is 11.
✦ Try This: Let Sₙ denote the sum of the first n terms of an AP. 3Sₙ = S₂ₙ. what is 3Sₙ:S₂ₙ
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 33
The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n
Summary:
The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. The value of n is 11.
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