How many terms of Arithmetic Progression 9,17,25---are to be taken that their sum is 636 ?

Question

Answer

Answer

A=9
d=17-9=8
sn=n/2(2a+(n-1)d)
636=n/2(2(9)+(n-1)8)
4n^2+5n-636=0
4n^2+53n-48n-636=0
n-12=0
n=12
or n= -53/4 which is not possible
so n=12

Answer

12 terms

Answer

Given:
Arithmetic progression: 9, 17, 25...

To find:
The number of terms required to sum up to 636.

Explanation:
To solve this problem, we need to find the number of terms in the given arithmetic progression that sum up to 636. Let's break down the problem step by step.

Step 1: Identify the common difference:
In an arithmetic progression, the difference between any two consecutive terms is constant. In this case, we can observe that the common difference is 17 - 9 = 8.

Step 2: Identify the last term:
Since we don't know the number of terms, let's assume the last term of the arithmetic progression is represented by 'a'. To find 'a', we can use the formula:
a = first term + (n-1) * common difference
where 'n' represents the number of terms. Substituting the given values, we get:
a = 9 + (n-1) * 8

Step 3: Find the sum of the arithmetic progression:
The sum of an arithmetic progression can be calculated using the formula:
Sum = (n/2) * (first term + last term)
Substituting the given values, we get:
636 = (n/2) * (9 + a)

Step 4: Solve the equation:
Substituting the value of 'a' from Step 2 into the equation from Step 3, we get:
636 = (n/2) * (9 + 9 + (n-1) * 8)
Simplifying further:
636 = (n/2) * (18 + 8n - 8)
636 = (n/2) * (10 + 8n)
1272 = n(10 + 8n)
8n^2 + 10n - 1272 = 0

Step 5: Solve the quadratic equation:
We can solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we can factor the equation as follows:
(n - 12)(8n + 106) = 0
n - 12 = 0 or 8n + 106 = 0
n = 12 or n = -106/8 (ignoring the negative value)

Step 6: Determine the number of terms:
Since the number of terms cannot be negative, we consider the positive value n = 12. Therefore, the number of terms required to sum up to 636 is 12.

Answer:
The number of terms required to sum up to 636 in the arithmetic progression 9, 17, 25... is 12.

Similar Class 10 Doubts

How many term of arithmetic progression,9, 17,25.are to be taken so their sum is 636?

how many terms of the AP:9,17,25,... must be taken to give a sum of 636 Related: Exercise 1 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics?

The first, second and last term of an arithmetic progression are respectively 4, 7 and 31. How many terms are there in the given arithmetic progression?

The first three terms of an AP are 3y-1,3y+5and 5y+1 respectively.find the value of y Related: Introduction - Arithmetic Progression (Hindi) , Maths, Class 10

There are n arithmetic mean between 11 and 53.each of them are integer. How many district arithmetic progression possible from the above data?

Top Courses for Class 10View all

Science Class 10

Social Studies (SST) Class 10

Mathematics (Maths) Class 10

English Grammar Advanced

English Grammar Basic

Top Courses for Class 10

Top Courses for Class 10

Suggested Free Tests

Related Content

About Class 10

Class 10 Study Material

Class 10 All Subjects

NCERT Textbook Solutions

Other Classes

Important Links

Company