If *S*_{n} denotes the sum of first *n* terms of an A.P., prove that *S*_{30} = 3[*S*_{20} − *S*_{10}]

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#### Solution

We know

S_{n}=`n/2`[2a+(n−1)d]

⇒S_{20}=`20/2`[2a+(20−1)d] and S_{10}=`10/2`[2a+(10−1)d]

⇒S_{20}=10[2a+19d] and S_{10}=5[2a+9d]

⇒S_{20}=20a+190d and S_{10}=10a+45d

3(S_{20}−S_{10})=3(20a+190d−10a−45d)

= 3(10a+145d)

= 15(2a+29d)

=`30/2`[2a+(30−1)d]

= S_{30}

∴ S_{30}=3(S_{20}−S_{10)}

Concept: Sum of First n Terms of an AP

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