Question

if Sr denotes the sum of first r terms of an A.P.,then find S3r-Sr-1/S2r-S2r-1

Answer 1

Let a is the first term and d is the common difference of AP
∵ Sn = n/2{2a + (n - 1)d}
and $\bold{S_n-S_{n-1}=T_n}$
so, $\bold{S_{2r} - S_{2r-1}=T_{2r}}$
And $\bold{S_{3r}-S_{r-1}=\frac{3r}{2}[2a+(3r-1)d]-\frac{r-1}{2}[2a+(r-2)d]}\\\\=\bold{2a[3r/2 - (r-1)/2] + d/2[(9r^2 - 3r) - (r^2-3r+2)]} \\\\=\bold{2a\frac{2r+1}{2}+d(4r^2-1)} \\\\=\bold{(2r+1)[2a+(2r-1)d]}\\\\\bold{=(2r+1).T_{2r}}$

Hence, $\bold{\frac{S_{3r}-S_{r-1}}{S_{2r}-S_{2r-1}}=\frac{(2r+1).T_{2r}}{T_{2r}}}$
= (2r + 1)