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Let `S_n` denote the sum of first n terms of an AP and `3S_n=S_(2n)`What is `S_(3n):S_n` equal to?What is `S_(3n):S_(2n)` equal to? |
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Answer» `S_n=n/2[2a+(n-1)d]` `3S_n=S_(2n)` `3(n/2)[2a+(n-1)d]=(2n)/2[2a+(2n-1)d]` `3/2[2a+(n-1)d]=2a+(2n-1)d` `3a+3/2nd-3/2d=2a+2nd-d` `a=(nd)/2+d/2` `a=(n+1)d/2` `2a=(n+1)d` `S_(3n)/S_n=((3n/2)[2a+(3n-1)d])/(n/2[2a+(n-1)d]` `=(3[nd+d+3nd-d])/(nd+d+nd-d)` `(12nd)/(2nd)` `S_(3n):S_n=6:1` option b is correcct. |
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