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B
1230.12
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C
1230.21
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D
1230.21
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Solution
The correct option is D1230.21 Using a0=1 and a−m=1am, (1×103)+(2×102)+(3×100)+(1×10−1)+(2×10−2) =1000+200+3+(110)+(2100) =1000+200+3+0.1+0.02 =1203.12