A) \[{{S}_{2}}=3{{S}_{3}}-2{{S}_{1}}\]
B) \[{{S}_{3}}=4({{S}_{1}}+{{S}_{2}})\]
C) \[{{S}_{3}}=3({{S}_{2}}-{{S}_{1}})\]
D) \[{{S}_{3}}=2({{S}_{2}}+{{S}_{1}})\]
Correct Answer: C
Solution :
Let the first term and common difference of \[\text{a}\] \[AP\] be \[a\] and \[d\] respectively. \[\therefore \] \[{{S}_{1}}=\frac{n}{2}[2a+(n-1)d]\] \[{{S}_{2}}=\frac{2n}{2}[2a+(2n-1)d]\] and \[{{S}_{3}}=\frac{3n}{2}[2a+(3n-1)d]\] Now, \[{{S}_{2}}-{{S}_{1}}=\frac{n}{2}[2a+(3n-1)d]\] \[=\frac{{{S}_{3}}}{3}\] \[\Rightarrow \] \[3({{S}_{2}}-{{S}_{1}})={{S}_{3}}\]
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