Advertisements
Advertisements
प्रश्न
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 2 `1/2` m apart, what is the length of the wood required for the rungs?
[Hint: number of rungs = `250/25+ 1`]

Advertisements
उत्तर
It is given that the rungs are 25 cm apart and the top and bottom rungs are `2 1/2` m
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
Sn =` n/2(a+l)`
∴ S10 = `11/2(45+25`)
= `11/2 (70)`
= 385 cm
Therefore, the length of the wood required for the rungs is 385 cm.
APPEARS IN
संबंधित प्रश्न
How many terms of the A.P. 18, 16, 14, .... be taken so that their sum is zero?
In an AP, given a = 7, a13 = 35, find d and S13.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees, and so on till class XII. There are three sections of each class. How many trees will be planted by the students?
Find the sum of the following arithmetic progressions:
3, 9/2, 6, 15/2, ... to 25 terms
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
If 4 times the 4th term of an A.P. is equal to 18 times its 18th term, then find its 22nd term.
If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero
If an denotes the nth term of the AP 2, 7, 12, 17, … find the value of (a30 - a20 ).
How many terms of the AP 63, 60, 57, 54, ….. must be taken so that their sum is 693? Explain the double answer.
Choose the correct alternative answer for the following question .
In an A.P. first two terms are –3, 4 then 21st term is ...
Q.2
Q.16
Find the sum of the first 10 multiples of 6.
Find whether 55 is a term of the A.P. 7, 10, 13,... or not. If yes, find which term is it.
If the sum of first n terms of an AP is An + Bn² where A and B are constants. The common difference of AP will be ______.
How many terms of the AP: –15, –13, –11,... are needed to make the sum –55? Explain the reason for double answer.
Kanika was given her pocket money on Jan 1st, 2008. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.
The nth term of an A.P. is 6n + 4. The sum of its first 2 terms is ______.
