The 4th term from the end of the AP: -11, -8, -5, ..., 49 is
a. 37
b. 40
c. 43
d. 58

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.

The nth term of an AP from the end is

aₙ = l -(n - 1)d------------------------(1)

a = first term

aₙ = nth term

d = common difference.

l = Last term.

From the question,

l = 49

d = -8 - (-11) = -8 + 11 = 3

From(1),we get,

a₄ = 49 - (4 - 1) 3

a₄ = 49 - 3(3)

a₄ = 49 - 9

a₄ = 40.

Therefore, the 4th term is 40.

✦ Try This: The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.1 Problem 13

The 4th term from the end of the AP: -11, -8, -5, ...,49 is, a. 37, b. 40, c. 43, d. 58

Summary:

An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. The 4th term from the end of the AP: -11, -8, -5, ..49 is 40

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