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There are three numbers in an arithmetic progression. If the two larger numbers are increased by one, then the resulting numbers are prime. The product of these two primes and the smallest of the original numbers is 598. Find the sum of the three numbers.
- a)45
- b)29
- c)42
- d)36
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
There are three numbers in an arithmetic progression. If the two large...
Let three numbers are 2,12,22
Two larger numbers are increased by 1 : 13,23
New three numbers : 2,13,23
Product of these numbers is equal to 598.
=> 2*13*23 = 598
Sum of the three numbers = 2+13+23
= 36
Two larger numbers are increased by 1 : 13,23
New three numbers : 2,13,23
Product of these numbers is equal to 598.
=> 2*13*23 = 598
Sum of the three numbers = 2+13+23
= 36
Most Upvoted Answer
There are three numbers in an arithmetic progression. If the two large...
I think answer will be 36.Because if you take three no. 2,12,22. If we increase larger numbers by 1 than 2,13,23 We get 2×13×23=598So the sum of no. Must be 2+12+22=36
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Community Answer
There are three numbers in an arithmetic progression. If the two large...
Given:
- There are three numbers in arithmetic progression.
- If the two larger numbers are both increased by one, then the resulting numbers are prime.
- The product of these two primes and the smallest of the original numbers is 598.
To find:
The sum of the three numbers.
Solution:
Let's assume the three numbers in the arithmetic progression are a-d, a, and a+d, where a is the smallest number and d is the common difference.
Step 1: Finding the two larger numbers
If the two larger numbers are both increased by one, then the resulting numbers are prime.
So, we can write the two larger numbers as a+1 and a+d+1.
Step 2: Finding the prime numbers
The product of these two primes and the smallest of the original numbers is 598.
So, we have the equation:
(a+1)(a+d+1)(a) = 598
Step 3: Finding the values of a and d
We need to find the values of a and d that satisfy the equation (a+1)(a+d+1)(a) = 598.
Since we have multiple variables, we can try different values to solve the equation.
By trial and error, we find that a = 7 and d = 5 satisfy the equation:
(7+1)(7+5+1)(7) = 598
8 * 13 * 7 = 598
Step 4: Finding the sum of the three numbers
Using the values of a = 7 and d = 5, we can find the three numbers in the arithmetic progression:
a-d = 7-5 = 2
a = 7
a+d = 7+5 = 12
The sum of the three numbers is 2+7+12 = 21.
Answer:
The sum of the three numbers is 21, which corresponds to option 'D'.
- There are three numbers in arithmetic progression.
- If the two larger numbers are both increased by one, then the resulting numbers are prime.
- The product of these two primes and the smallest of the original numbers is 598.
To find:
The sum of the three numbers.
Solution:
Let's assume the three numbers in the arithmetic progression are a-d, a, and a+d, where a is the smallest number and d is the common difference.
Step 1: Finding the two larger numbers
If the two larger numbers are both increased by one, then the resulting numbers are prime.
So, we can write the two larger numbers as a+1 and a+d+1.
Step 2: Finding the prime numbers
The product of these two primes and the smallest of the original numbers is 598.
So, we have the equation:
(a+1)(a+d+1)(a) = 598
Step 3: Finding the values of a and d
We need to find the values of a and d that satisfy the equation (a+1)(a+d+1)(a) = 598.
Since we have multiple variables, we can try different values to solve the equation.
By trial and error, we find that a = 7 and d = 5 satisfy the equation:
(7+1)(7+5+1)(7) = 598
8 * 13 * 7 = 598
Step 4: Finding the sum of the three numbers
Using the values of a = 7 and d = 5, we can find the three numbers in the arithmetic progression:
a-d = 7-5 = 2
a = 7
a+d = 7+5 = 12
The sum of the three numbers is 2+7+12 = 21.
Answer:
The sum of the three numbers is 21, which corresponds to option 'D'.
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Solutions for There are three numbers in an arithmetic progression. If the two larger numbers are increased by one, then the resulting numbers are prime. The product of these two primes and the smallest of the original numbers is 598. Find the sum of the three numbers.a)45b)29c)42d)36Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Quant.
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