A) 47
B) 42
C) 52
D) 57
Correct Answer: C
Solution :
\[S={{a}_{1}}+{{a}_{2}}+......{{a}_{30}}~\,\,\Rightarrow \,\,S\,\,=\text{ }\frac{30}{2}\left( {{a}_{1}}+{{a}_{30}} \right)\] \[T\,\,=\,\,{{a}_{1}}\,\,+\,\,{{a}_{2}}\,\,+\,{{a}_{5}}\,+.....\,\,{{a}_{29}}\,\,\Rightarrow \,\,T\,\,=\,\,\frac{15}{2}[{{a}_{1}}\,+\,\,{{a}_{29}}]\] \[\Rightarrow \,\,\,2T\,\,=\,\,15\,\,({{a}_{1}}\,+\,\,{{a}_{29}})\] \[S-\text{ }27\text{ }=\text{ }75\] \[15\left( {{a}_{1}}+{{a}_{30}} \right)-15\left( {{a}_{1}}+{{a}_{29}} \right)=\text{ }75\] \[{{a}_{30}}\,-\,{{a}_{29}}\,\,=\,\,5\] \[d\,\,=\,\,5\] \[{{a}_{5}}=\text{ }27\] \[{{a}_{1}}+4d=\text{ }27\] \[{{a}_{1}}+\text{ }20\text{ }=\text{ }27\] \[{{a}_{1}}=7\] \[{{a}_{10}}={{a}_{1}}+9d\] \[=\text{ }7\text{ }+\text{ }9\left( 5 \right)\] \[=7\,\,+\,\,45\] \[=\,\,52\]
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