A) 21
B) 15
C) 16
D) 19
Correct Answer: B
Solution :
[b] \[{{S}_{3n}}=\frac{3n}{2}[2a+(3n-1)d]\] \[{{S}_{n-1}}=\frac{n-1}{2}[2a+(n-2)d]\] \[\Rightarrow {{S}_{3n}}-{{S}_{n-1}}=\frac{1}{2}[2a(3n-n+1)]\] \[+\frac{d}{2}[3n(3n-1)-(n-1)(n-2)]\] \[=\frac{1}{2}[2a(2n+1)+d(8{{n}^{2}}-2)]\] \[=a(2n+1)+d(4{{n}^{2}}-1)\] \[=(2n+1)[a+(2n-1)d]\] \[{{S}_{2n}}-{{S}_{2n-1}}={{T}_{2n}}=a+(2n-1)d\] \[\Rightarrow \frac{{{S}_{3n}}-{{S}_{n-1}}}{{{S}_{2n}}-{{S}_{2n-1}}}=(2n+1)\] Given, \[\Rightarrow \frac{{{S}_{3n}}-{{S}_{n-1}}}{{{S}_{2n}}-{{S}_{2n-1}}}=31\Rightarrow n=15\]
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