Answer:
Given, \[{{S}_{n}}=\frac{1}{2}(3{{n}^{2}}+7n)\] Now, \[{{S}_{1}}=\frac{1}{2}[3{{(1)}^{2}}+7(1)]=5=a\] (First term) And, \[{{S}_{2}}=\frac{1}{2}[3{{(2)}^{2}}+7(2)]=13\] Second term \[({{a}_{2}})=13-5=8\] \[\therefore a=5,d=3\] We know, \[{{T}_{n}}=a+(n-1)d\] \[=5+(n-1)(3)\] \[=5+3n-3\] \[\therefore {{T}_{n}}=3n+2\] And \[{{T}_{20}}=5+(20-1)3\] \[=5+19\times 3\] \[\therefore {{T}_{20}}=62\]
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