Answer:
Let a be the first term and d be the common difference of the A.P. \[\because {{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\] \[\therefore {{S}_{30}}=\frac{30}{2}[2a+(30-1)d]\] \[=15(2a+29d)\] \[=30a+435d)\] \[{{S}_{20}}=\frac{20}{2}[2a+(20-1)d]\] \[=10[2a+19d]\] \[=20a+190d\] And, \[{{S}_{10}}=\frac{10}{2}[2a+(10-1)d]\] \[=5[2a+9d]\] \[=10a+45d\] Now, \[3[{{S}_{20}}-{{S}_{10}}]\]\[=3[20a+190d-10a-45d]\] \[=3[10a+145d]\] \[=30a+435d\] \[={{S}_{30}}\] \[\therefore {{S}_{30}}=3[{{S}_{20}}-{{S}_{10}}]\] Hence Proved.
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