Question

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

Answer 1

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Answer 2

Answer:

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)

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