|
|
|
|
|
|
|
|
|
|
Method and examples
|
|
|
|
Arithmetic Progression |
|
|
|
|
|
|
Problem 16 of 19 |
|
|
|
16. If Sn is sum of n even terms of arithmetic progression series and Sn' is sum of n odd terms of arithmetic progression series then prove that Sn = (1 + 1/n) Sn'
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Solution
|
Solution provided by AtoZmath.com
|
|
|
This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)
| Arithmetic Progression |
16. If Sn is sum of n even terms of arithmetic progression series and Sn' is sum of n odd terms of arithmetic progression series then prove that Sn = (1 + 1/n) Sn'
Here `S_n = 2 + 4 + 6 + ... + 2n`
`:. S_n = n/2 [ 2a + (n - 1) d ]`
`= n/2 [ 2(2) + (n - 1) * 2 ]` (because a = 2 and d = 2)
`= n/2 [ 4 + 2n - 2 ]`
`= n/2 [ 2n + 2 ]`
`= n ( n + 1 ) `
Now, `S_(n') = 1 + 3 + 5 + ... + (2n - 1)`
`:. S_(n') = n/2 [ 2a + (n - 1) d ]`
`= n/2 [ 2(1) + (n - 1) * 2 ]` (because a = 1 and d = 2)
`= n/2 [ 2 + 2n - 2 ]`
`= n/2 [ 2n ]`
`= n^2`
Now, `(S_n)/(S_n') = (n (n + 1))/(n^2) `
`:. (S_n)/(S_n') = ((n + 1))/n`
`:. S_n = (1 + 1/n) × S_(n')` (Proved)
|
|
|
|
|
|
|
Share this solution or page with your friends.
|
|
|
|
