If S, denotes the sum of first n terms of an AP prove that, S30 = 3(S20 - S10)
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Sn=n/2[2a+(n-1)d]
S30=30/2[2a+29d]
S30=15(2a+29d)
S30 =30a+435d
S20=20/2(2a+19d)
S20=10(2a+19d)
S20= 20a+190d
S10=10/2(2a+9d)
S10=5(2a+9d)
S10=10a+45d
3(S20-S10)=3(20a+190d-10a-45d)
=3(10a+145d)
=30a+435d
S30=30a+435d
Hence S30=3(S20-S10)
S30=30/2[2a+29d]
S30=15(2a+29d)
S30 =30a+435d
S20=20/2(2a+19d)
S20=10(2a+19d)
S20= 20a+190d
S10=10/2(2a+9d)
S10=5(2a+9d)
S10=10a+45d
3(S20-S10)=3(20a+190d-10a-45d)
=3(10a+145d)
=30a+435d
S30=30a+435d
Hence S30=3(S20-S10)
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