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Arithmetic Progressions

Question

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A ladder has rungs 25 cm apart. (see the given Fig.). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2 1 half straight m$ apart, what is the length of the wood required for the rungs?

Solution

Number of rungs (n) = $250 over 25 equals 10$
$open square brackets 2 1 half straight m equals 250 space cm close square brackets$

Hence, there are 10 rungs.
The length of the wood required for the rungs (S₁₀).

straight S subscript 11 equals 10 over 2 left parenthesis 45 plus 25 right parenthesis space equals space 350 space cm.

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 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

For the following APs, write the first term and the common difference
3, 1, – 1, – 3, . . .

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