Q. 204.0( 3 Votes )

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Answer :

Let a be the first term and d be the common difference of A.P.


We know that Sn =


…………….(1)


…………….(2)


…………….(3)


To prove: S30=3(S20-S10)


Solving RHS,


RHS=3(S20-S10)


=3[(20a + 190d)-(10a + 45d)] (From eq (2) and (3))


=3[10a + 145d]


=30a + 435d


From eq(1),


RHS=30a + 435d=S30=LHS


Hence proved.


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