Answer:
Given, \[a=5,{{a}_{n}}=45,{{S}_{n}}=400\] We have, \[{{S}_{n}}=\frac{n}{2}[a+{{a}_{n}}]\] \[\Rightarrow \] \[400=\frac{n}{2}[5+45]\] \[\Rightarrow \] \[400=\frac{n}{2}[50]\] \[\Rightarrow \] \[25n=400\Rightarrow n=\frac{400}{25}\] \[\Rightarrow \] \[n=16\] Now, \[{{a}_{n}}=a+(n-1)d\] \[\Rightarrow \] \[45=5+(16-1)d\] \[\Rightarrow \] \[45-5=15\,d\] \[\Rightarrow \] \[15\,d=40\] \[\Rightarrow \] \[d=\frac{8}{3}\] So, \[n=16\] and \[d=\frac{8}{3}\]
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