The sum of the first three numbers in an AP is 18. If the product of t...
the terms are either 2,6,10 or 15,6,-3.
Step-by-step explanation:
It is given that the sum of first three numbers in an AP is 18. Product of the first and the third term is 5 times the common difference
Let first three numbers in the AP are a-d, a, a+d.
Sum of these three terms is 18.
Divide both sides by 3
The value of a is 6.
The product of the first and the third term is 5 times the common difference.
If the common difference is 4, then
Therefore the first three terms are 2, 6 and 10.
If the common difference is -9, then
Therefore the first three terms are 15, 6 and -3.
This question is part of UPSC exam. View all Class 10 courses
The sum of the first three numbers in an AP is 18. If the product of t...
The sum of the first three numbers in an AP is 18. If the product of t...
Given Information:
- The sum of the first three numbers in an arithmetic progression (AP) is 18.
- The product of the first and the third term is 5 times the common difference.
Let's solve the problem step by step:
Step 1: Understanding Arithmetic Progression (AP)
An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is called the common difference (d).
Step 2: Setting up the problem
Let's assume the first term of the AP as 'a', the second term as 'a + d', and the third term as 'a + 2d'. Here, 'a' represents the first term, and 'd' represents the common difference.
Step 3: Sum of the first three terms
The sum of the first three terms of an AP can be calculated using the formula:
Sum = (n/2)(2a + (n-1)d),
where 'n' is the number of terms.
In this case, the sum of the first three terms is given as 18.
So, we can set up the equation as:
18 = (3/2)(2a + 2d)
Simplifying the equation, we get:
18 = 3a + 3d
Dividing by 3, we have:
6 = a + d --- (Equation 1)
Step 4: Product of the first and third term
The product of the first and the third term can be expressed as:
Product = (a)(a + 2d)
According to the given information, the product is 5 times the common difference:
(a)(a + 2d) = 5d
Step 5: Solving the equations
Now, we have two equations:
Equation 1: 6 = a + d
Equation 2: (a)(a + 2d) = 5d
We can solve these equations simultaneously to find the values of 'a' and 'd'.
Step 6: Solution
Using Equation 1, we can express 'a' in terms of 'd':
a = 6 - d
Substituting this value in Equation 2, we get:
(6 - d)(6 - d + 2d) = 5d
Simplifying the equation:
(6 - d)(6 + d) = 5d
36 - d^2 = 5d
Rearranging the equation:
d^2 + 5d - 36 = 0
Factorizing the quadratic equation:
(d + 9)(d - 4) = 0
So, d = -9 or d = 4
If we substitute d = -9 in Equation 1, we get a negative value for 'a', which is not possible as it represents the first term of the AP. Therefore, we disregard d = -9 and consider d = 4.
Substituting d = 4 in Equation 1, we get:
6 = a + 4
a = 2
Thus, the three numbers in the arithmetic progression are:
First term (a) = 2
Second term (a + d) =
To make sure you are not studying endlessly, EduRev has designed Class 10 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 10.