Find the value of the middle most term (s) of the AP : -11, -7, -3,..., 49
Solution:
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
From the question,
a = -11
d = -7 - (-11) = 4
aₙ = 49.
The nth term of the AP is
aₙ = a + (n - 1) d
Substituting the above values, we get,
49 = -11 + (n - 1) × 4
60 = (n - 1) × 4
n = 16.
n is an even number.
Hence, there will be two middle terms which are 16/2 th and (16/2 +1)th.
They are the 8th term and the 9th term.
a₈ = a + 7d = -11 + 7 × 4 = 17
a₈ = a + 8d = -11 + 8 × 4 = 21.
Therefore, the values of the two middle most terms are 17 and 21, respectively.
✦ Try This: If Sₙ, denotes the sum of first n terms of an A.P., prove that S₁₂ = 3(S₈ - S₄)
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.3 Sample Problem 2
Find the value of the middle most term (s) of the AP : -11, -7, -3,..., 49
Summary:
The value of the middle most term (s) of the AP : -11, -7, -3,..49 are 17 and 21
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