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8. Sequences and Series
medium
જો એક સમાંતર શ્રેણી માટે $S_{2n} = 2S_n$ હોય, તો $S_{3n}/ S_n = …….$
A
$4$
B
$6$
C
$8$
D
$10$
Solution
અહીં $S_{2n} = 2S_n$
$\therefore \,\,\frac{2n}{2}\,(2a\,+\,(2n\,-\,1)\,d)\,\,=\,\,3\,\frac{n}{2}\,(2a\,+\,(n\,-\,1)\,d)\,\,\,\,$
$\therefore \,\,2a\,+\,(2n\,-\,1)\,d\,\,=\,\,\frac{3}{2}\,\,(2a\,+\,(n\,-\,1)\,d)$
$\therefore \,\,(2a\,+\,2nd\,-\,d)\,\,3\,\,(2a\,+\,nd\,-\,d)\,$
$\therefore \,\,\,\,2a\,\,=\,\,d\,\,(n\,+\,1)$
હવે,$\frac{{{S}_{3n}}}{{{S}_{n}}}\,\,=\,\,\frac{\frac{3n}{2}\,\left[ 2a\,+\,(3n\,-\,1)\,d \right]}{\frac{n}{2}\,\left[ 2a\,+\,(n\,-\,1)\,d \right]}$
$=\,\,\frac{3\,\left[ d\,(n\,+\,1)\,+\,(3n\,-\,1)\,d \right]}{d\,(n\,+\,1)\,+\,(n\,-\,1)\,d}$
$=\,\,\frac{3d\,(4n)}{d\,(2n)}\,\,=\,\,6$
Standard 11
Mathematics
