1 if S1 is the sum of an A P of 'n' odd number of terms and S2 - Maths - Arithmetic Progressions

Question

1 if S1 is the sum of an A P of 'n' odd number of
terms and S2 the sum of the terms of the series in odd
places, then find S1/S2
2 if Sn denotes the sum of n terms of an A P with first term a
and common difference d such that Sx/Skx is independent of x,
then
a) d=a b) d=2a c) a=2d d) d=-a

Pronay · Accepted Answer

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)

1. if S1 is the sum of an A.P. of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then find S1/S2. 2. if Sn denotes the sum of n terms of an A.P. with first term a and common difference d such that Sx/Skx is independent of x, then a) d=a b) d=2a c) a=2d d) d=-a