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If `S_(n)` denotes the sum of first n terms of an A.P., then `(S_(3n)-S_(n-1))/(S_(2n)-S_(n-1))` is equal toA. 2n-1B. 2n+1C. 4n+1D. 2n+3 |
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Answer» Correct Answer - B We have, `(S_(3n)-S_(n-1))/(S_(2n)-S_(n-1))=((3n)/(2){2a+(3n-1)d}-(n-1)/(2){2a+(n-2)d})/((2n)/(2){2a+(2n-1)d}-(2n-1)/(2){2a+(2n-2)d})` `=(2a(3n-n+1)+{3n(3n-1)-(n-1)(n-2)}d)/(2a(2n-2n+1)+{2n(2n-1)-(2n-1)(2n-2)}d)` `=(2a(2n+1)+{9n^(2)-3n-n^(2)+3n-2}d)/(2a+{4n^(2)-2n-4n^(2)+6n-2}d)` `=(2a(2n+1)+(8n^(2)-2)d)/(2a+(4n-2)d)` `=(2(2n+1){a+(2n-1)d})/(2{a+(2n-1)d})=2n+1` |
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