if Sr denotes the sum of first r terms of an A.P.,then find S3r-Sr-1/S2r-S2r-1
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Let a is the first term and d is the common difference of AP
∵ Sn = n/2{2a + (n - 1)d}
and
so,
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}}](https://tex.z-dn.net/?f=%5Cbold%7BS_%7B3r%7D-S_%7Br-1%7D%3D%5Cfrac%7B3r%7D%7B2%7D%5B2a%2B%283r-1%29d%5D-%5Cfrac%7Br-1%7D%7B2%7D%5B2a%2B%28r-2%29d%5D%7D%5C%5C%5C%5C%3D%5Cbold%7B2a%5B3r%2F2+-+%28r-1%29%2F2%5D+%2B+d%2F2%5B%289r%5E2+-+3r%29+-+%28r%5E2-3r%2B2%29%5D%7D+%5C%5C%5C%5C%3D%5Cbold%7B2a%5Cfrac%7B2r%2B1%7D%7B2%7D%2Bd%284r%5E2-1%29%7D+%5C%5C%5C%5C%3D%5Cbold%7B%282r%2B1%29%5B2a%2B%282r-1%29d%5D%7D%5C%5C%5C%5C%5Cbold%7B%3D%282r%2B1%29.T_%7B2r%7D%7D++ "\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,
= (2r + 1)
∵ Sn = n/2{2a + (n - 1)d}
and
so,
And
Hence,
= (2r + 1)
duttaditya18p3zry7:
Please write a better answer. This one is weird and convoluted.
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