If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be
a. 7
b. 11
c. 18
d. 0

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 is

aₙ = a + (n - 1 )d.

a = first term

aₙ = nth term

d = common difference.

As per the question,

7a₇ = 11a₁₁

Since,aₙ = a + (n - 1 )d.

7[a + (7 - 1)d] = 11 [a + (11 - 1)d]

7(a + 6d) = 11 (a + 10d)

7a + 42d = 11a + 110d

42d -110d = 11a - 7a

4a + 68 d = 0

2(2a + 34d) = 0

2a + 34d = 0

a + 17d = 0----------------------(1)

18th term of an AP is

a₁₈ = a + (18 - 1)d

a₁₈ = a + 17d

a₁₈ = 0.

Therefore, the a₁₈ =0.

✦ Try This: Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term

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

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NCERT Exemplar Class 10 Maths Exercise 5.1 Problem 12

If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be, a. 7, b. 11, c. 18, d. 0

Summary:

If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be 0.

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