Find the sum of the two middle most terms of the AP: -4/3, -1, -2/3,.........,4 ⅓
Solution:
Given, the arithmetic series is -4/3, -1, -2/3,-------------,4 ⅓
We have to find the sum of the two middle most terms of the series.
From the given series,
First term, a = -4/3,
aₙ = 4 ⅓ = 13/3
Common difference, d = -1 - (-4/3)
= -1 + 4/3
= (-3+4)/3
d = 1/3
The n-th term of an AP is given by
aₙ = a + (n - 1)d
To find n,
13/3 = -4/3 + (n - 1)(1/3)
13/3 + 4/3 = (n - 1)(1/3)
17/3 = (1/3)(n - 1)
n - 1 = 17
n = 17 + 1
n = 18
The two middle most terms of the 18 terms of the series are 9th and 10th terms.
So, a₉ = -4/3 + (9 - 1)(1/3)
= -4/3 + (8/3)
= (-4+8)/3
a₉ = 4/3
a₁₀ = -4/3 + (10 - 1)(1/3)
= -4/3 + (9/3)
= (-4+9)/3
a₁₀ = 5/3
Now, sum of two middle terms = a₉ + a₁₀
= 4/3 + 5/3
= (4+5)/3
= 9/3
= 3
Therefore, the sum of the two middlemost terms of the given AP is 3.
✦ Try This: Find the sum of the two middlemost terms of the AP: 2, 4, 6,..........,24
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 19
Find the sum of the two middlemost terms of the AP: -4/3, -1, -2/3,.........,4 ⅓
Summary:
An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same. The sum of the two middle most terms of the AP -4/3, -1, -2/3,........,4 ⅓ is 3.
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