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Question
If Sn denotes the sum of first n terms of an A.P., prove that S30 = 3[S20 − S10]
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Solution
We know
Sn=`n/2`[2a+(n−1)d]
⇒S20=`20/2`[2a+(20−1)d] and S10=`10/2`[2a+(10−1)d]
⇒S20=10[2a+19d] and S10=5[2a+9d]
⇒S20=20a+190d and S10=10a+45d
3(S20−S10)=3(20a+190d−10a−45d)
= 3(10a+145d)
= 15(2a+29d)
=`30/2`[2a+(30−1)d]
= S30
∴ S30=3(S20−S10)
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