Divide 32 into 4 parts such that all are in A.P such that the ratio of product of extremes to the p
Asked by | 23rd Feb, 2008, 06:33: PM
Your question was
Divide 32 into 4 parts such that all are in A.P such that the ratio of product of extremes to the product of means is 7: 15.
Let us take the 4 numbers in AP as a-3d,a-d,a+d,a+3d
Now a-3d+a-d+a+d+a+3d = 32
So 4a = 32
a = 8
Also According to II condition
(a2 -9d2):(a2-d2) =7:15
So 15( a2 -9d2) = 7(a2-d2)
15(64 - 9d2) = 7(64-d2)
Solving we get d = 2
So the AP is 2,6,10,14 or 14,10,6,2
Answered by | 5th Dec, 2017, 06:01: PM
- Eg4 please...Why are 'means' taken as a-d,a+d and not the other two?
- What is the common difference of the arithmetic sequence 17,15,13.......
- Which of the following are A.P's? If they form an AP, find the common difference d. (1) 4, 0, -4, -8, -12, ... (2) -2, 0, 2, 5, 8, 10, ...
- Write the next two terms of the following AP: ……….
- Write first term and common difference for the following AP'S
- Write the four terms of the AP when first term is and the common difference is ?
- Write the next two terms of the following AP: 3, 1, -1, -3, ……..?
- Find the terms of the sequence defined by
- Write the next two terms of the following sequence
- Write the first three terms of sequences starting the value of n from 1.
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