Question 128
If $$S_1, S_2$$ and $$S_3$$ denote the sums of the first n, 2n and 3n terms respectively of an arithmetic progression then $$\frac{S_3}{S_2 - S_1} =$$
Solution
$$S_1=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$$
$$S_2=n\left[2a+\left(2n-1\right)d\right]$$
$$S_3=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]$$
The desired operation results in:
$$\frac{3\left[2a+\left(3n-1\right)d\right]}{2\left[2a+\left(2n-1\right)d\right]-2a-\left(n-1\right)d}$$
=$$\frac{\left[6a+9nd-3d\right]}{2a+3nd-d}=3$$
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