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(1) For an A.P., if S10 150 and S 126, 10(A) 24 (B) 22 (C) 26 (D) 20 |
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Answer» 150=5(2a+9d)30=2a+9d 126=(9/2)[2a+8d]28=2a+8d Subtract2=d30-18=2aa=12/2=6 t10=a+9d6+18=24Option A |
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