Let a1,a2,......,a30 be an A.P., S=30∑i=1ai and T=15∑i=1a2i−1. If a5=27 and S−2T=75, then a10 is equal to :
A
42
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B
47
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C
57
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D
52
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Solution
The correct option is D 52 Given : a1,a2,......,a30 is an A.P. S=30∑i=1ai T=15∑i=1a2i−1 a5=27 and S−2T=75 Now assuming the first term as 'a' and the common difference to be 'd', S=a1+a2+a3+.......+a30 T=a1+a3+a5+.......+a29
Writing S−2T and rearranging the terms, we will get S−2T=(a2−a1)+(a4−a3)+(a6−a5)+.......(a30−a29)=75 ⇒15×d=75⇒d=5 As given, a5=27⇒a+4.d=27 ⇒a=7 ∴a10=a+9d=7+9×5=52