Question

If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then S1/S2 =

Answer 1

1) S1: = 1 + 3 + 5 + 7 + ---- + (2n-1) [nth term; a + (n-1)d; 
==> nth term = 1 + (n-1)2 = (2n-1)] 

2) Sum to n terms of an AP: (n/2){1st term + Last term}; 
==> S1 = (n/2)*(1 + 2n -1) = (n^2) 

3) S2 = 1 + 5 + 9 + 13 + ---- + (2n-3) 
[Here a = 1; d = 4; no. of terms = n/2, assuming in S1, number of terms are even] 
==> S2 = {(n/2)/2}*(1 + 2n -3) = (n/4)*(2n-2) = (n)(n-1)/2 

So, S1/S2 = (n^2)/{(n)(n-1)/2} = 2n/(n-1) 

Now let us consider, the number of terms in S1 are odd; 
Then number of terms in S2 would be = (n+1)/2 
==> Last term here = 1 + {(n+1)/2 - 1}4 = 1 + 2n + 2 - 4 = 2n - 1 

==> S2 = {(n+1)/2)/2}*(1 + 2n -1) = (n)*(n+1)/2 

So, in this case the ratio S1/S2 = 2n/(n+1)

Hope This Helps :)

Answer 2

Answer:

S1/S2 = 2n/(n+1)

Step-by-step explanation:

1) S1: = 1 + 3 + 5 + 7 + ---- + (2n-1) [nth term; a + (n-1)d; 

==> nth term = 1 + (n-1)2 = (2n-1)] 

2) Sum to n terms of an AP: (n/2){1st term + Last term}; 

==> S1 = (n/2)*(1 + 2n -1) = (n^2) 

3) S2 = 1 + 5 + 9 + 13 + ---- + (2n-3) 

[Here a = 1; d = 4; no. of terms = n/2, assuming in S1, number of terms are even] 

==> S2 = {(n/2)/2}*(1 + 2n -3) = (n/4)*(2n-2) = (n)(n-1)/2 

So, S1/S2 = (n^2)/{(n)(n-1)/2} = 2n/(n-1) 

Now let us consider, the number of terms in S1 are odd; 

Then number of terms in S2 would be = (n+1)/2 

==> Last term here = 1 + {(n+1)/2 - 1}4 = 1 + 2n + 2 - 4 = 2n - 1 

==> S2 = {(n+1)/2)/2}*(1 + 2n -1) = (n)*(n+1)/2 

So, in this case the ratio S1/S2 = 2n/(n+1)