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Arithmetic progression

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Question

 Let Sn denote the sum of first n-terms of an A.P. If S4=16,S6=48 thenS10is

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Solution

S4=22a+3d=16  Sn=n22a+n-1d

2a+3d=81S6=3[2a+5d]=482a+5d=16221d=122a=44a=22S10=102[2a+9d]=5[44+9(12)]=5(44108)=320, S10=320


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