A) 57
B) 47
C) 42
D) 52
Correct Answer: D
Solution :
\[{{a}_{1}},\]\[{{a}_{2}},\] ?. \[{{a}_{30}},\] A.P., let d be common difference ?d? |
\[\text{S}=\sum\limits_{i=l}^{30}{{{a}_{i}}},\] \[\text{T}=\sum\limits_{i=l}^{18}{{{a}_{2i-\,1}}}\] |
\[{{a}_{5}}=27\] |
\[\Rightarrow \] \[{{a}_{1}}+4d=27\] |
\[\Rightarrow \] \[\text{S}\,-\,2\text{T}=75\] |
\[({{a}_{2}}-{{a}_{1}})+({{a}_{4}}-{{a}_{3}})\,+...+({{a}_{30}}-{{a}_{29}})=75\] |
\[15d=75\] |
\[d=5\] |
by (1), \[{{a}_{1}}=5\] |
\[\therefore \] \[{{a}_{10}}={{a}_{1}}+9d=7+45=52.\] |
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