Three numbers are in the ratio 3:7:9. If 5 is subtracted from the second, the resulting numbers are in AP. Find the original numbers.
Asked by Topperlearning User | 22nd Sep, 2017, 09:01: AM
Let the three numbers which are in the ratio 3:7:9 be 3x, 7x and 9x.
Now, on subtracting 5 from the second number, we get the three numbers as: 3x, 7x - 5 and 9x.
Since, these numbers are in AP, so we have:
2(7x - 5) = 3x + 9x
14x - 10 = 12x
2x = 10
x = 5
Hence, the original three numbers are 3x = 15, 7x = 35 and 9x = 45.
Answered by | 22nd Sep, 2017, 11:01: AM
- in an AP of 50terms the sum of first 10 term is 210 and sum of last 15 term is 2565 find AP
- If the sum of m terms of an AP is equal to the sum of either the next n terms or the next p terms of the same AP prove that (m+ n) [(1/m)-(1/p)] = (m + p) [(1/m) -(1/n)] .
- Find the sum of n terms of the sequence 5+55+555....
- The sum of three no.s in an AP=-6 andtheir product is 64, find the no.s
- find the sum of first 15 term of an A.P whose nth term is 3-2n?
- Q no 9
- Q2 Please answer the question in short cut but fundementally and step wize . I hardly able to understand your answer .
- Find the sum of all 3-digit numbers which leave remainder 3 when divided by 5.
- 27. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.
- Sum of the first n terms of AP is 3n^2/2 + 13n/2. Find the nth term of the AP (Substitution method not allowed)
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number