A) \[{{S}_{3}}=2\,\,({{S}_{1}}+{{S}_{2}})\]
B) \[{{S}_{3}}={{S}_{1}}+{{S}_{2}}\]
C) \[{{S}_{3}}=3\,({{S}_{2}}-{{S}_{1}})\]
D) \[{{S}_{3}}=3\,({{S}_{2}}+{{S}_{1}})\]
Correct Answer: C
Solution :
Given, \[{{S}_{1}}=\]Sum of n terms of an AP \[{{S}_{1}}=\frac{n}{2}\,\,[2a+(n-1)d]\] \[S{{ & }_{2}}=\] sum of 2n terms of an AP \[S{{ & }_{2}}=\frac{2n}{2}\,[2a+(2n-1)d]\] and \[{{S}_{3}}\] = Sum of 3n terms of an AP \[{{S}_{3}}=\frac{3n}{2}[2a+(3n-1)d]\] Now, \[{{S}_{2}}-{{S}_{1}}=\frac{n}{2}\,\,[4a+(4n-2)d-2a-(n-1)d]\] \[=n/2\,\,[2a+(3n-1)d]\] \[3({{S}_{2}}-{{S}_{1}})=\frac{3n}{2}[2u+(3n-1)d]\] \[3({{S}_{2}}-{{S}_{1}})={{S}_{3}}\]
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