NCERT MATHS CLASS - X
Arithmetic Progressions
Exercise 5.4
Question 3:
A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are m apart, what is the length of the wood required for the rungs?
∴ Total number of rungs
Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.
The length of the wood required for the rungs equals the sum of all the terms of this A.P.
First term, a = 45
Last term, l = 25
n = 11
Therefore, the length of the wood required for the rungs is 385 cm.
Question 4:
The houses of a row are number consecutively from 1 to 49. Show that there is a value of x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it. Find this value of x.
The number of houses are 1, 2, 3 … 49 Since the number of houses are in an A.P.with a=1 and d =1.
Let us consider that the number of 10th house is like this.
Sum of n terms in an A.P.
Sum of number of houses preceding 10th house = Sx − 1
Sum of number of houses following 10th house = S49 − Sx
It is given that these sums are equal to each other.
However, the house numbers are positive integers.
The value of x will be 35 only.
Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.
Previous Q.1,2