A) 2489
B) 4735
C) 2317
D) 2632
Correct Answer: D
Solution :
Let \[S=1+2+3+...........+100\] \[=\frac{100}{2}(1+100)=50(101)=5050\] Let \[{{S}_{1}}=3+6+9+12+.........+99\] =\[3(1+2+3+4+.........+33)\] =\[3.\frac{33}{2}(1+33)=99\times 17=1683\] Let \[{{S}_{2}}=5+10+15+........+100\] =\[5(1+2+3+........+20)\] =\[5.\frac{20}{2}(1+20)=50\times 21=1050\] Let \[{{S}_{3}}=15+30+45+........+90\] = \[15(1+2+3+........+6)\] = \[15.\frac{6}{2}(1+6)=45\times 7=315\] \ Required sum =\[S-{{S}_{1}}-{{S}_{2}}+{{S}_{3}}\] =\[5050-1683-1050+315\]= 2632.
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