A) 240
B) 120
C) 300
D) 180
E) 480
Correct Answer: A
Solution :
Let common difference\[{{d}_{1}}=-3\]and first term be a. \[\therefore \]Series become \[a,\text{ }a-3,\text{ }a-6,...,\text{ }a-27\] \[\therefore \] \[S=10a+(-3-6-....-27)\] \[\Rightarrow \] \[-30=10a-3(1+2+...+9)\] \[\Rightarrow \] \[-30=10a-3\left[ \frac{9(9+1)}{2} \right]\] \[\Rightarrow \] \[-30=10a-135\] \[\Rightarrow \] \[10a=105\] \[\Rightarrow \] \[a=\frac{105}{10}\] Now, correct common difference\[{{d}_{2}}=3\] \[\therefore \]Required sum \[=\frac{10}{2}\left[ 2\times \frac{105}{10}+(10-1)3 \right]\] \[=5\left[ \frac{105}{5}+27 \right]\] \[=5\times 48=240\]You need to login to perform this action.
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