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Arithmetic Progressions

Question

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The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

Solution

We have given that
4th term of an A.P.= a₄ = 0
∴ a + (4 – 1)d = 0
∴ a + 3d = 0
∴ a = –3d ….(1)

25th term of an A.P. = a₂₅= a + (25 – 1)d
= –3d + 24d ….[From the equation (1)]
= 21d

3 times 11th term of an A.P. = 3a₁₁= 3[a + (11 – 1)d]
= 3[a + 10d]
= 3[–3d + 10d]
= 3 × 7d
= 21d

∴ a₂₅ = 3a₁₁

i.e., the 25th term of the A.P. is three times its 11th term.

Some More Questions From Arithmetic Progressions Chapter

In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 50 for each succeeding metre.

In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.

In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 50 for each succeeding metre.

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a – 10, d = 10

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = –2, d = 0

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = 4, d = – 3

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

a = – 1.25, d = – 0.25

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