Answer:
In the given A.P., let first term = a and common difference = d Then, \[{{T}_{n}}=a+(n-1)d\] \[\Rightarrow {{T}_{13}}=a+(13-1)d=a+12d\] and \[{{T}_{3}}=a+(3-1)d=a+2d\] Now, \[{{T}_{13}}=4{{T}_{3}}\] (Given) \[a+12d=4(a+2d)\] \[a+12d=4a+8d\] \[3a=4d\] \[a=\frac{4}{3}d\] ?(i) Also, \[{{T}_{5}}=a+(5-1)d\] \[\Rightarrow a+4d=16\] ?(ii) Putting the value of a from eq. (i) in (ii), we get \[\frac{4}{3}d+4d=16\] \[4d+12d=48\] \[16d=48\] \[d=3\] Substituting \[d=3\] in eq. (ii), we get \[a+4(3)=16\] \[a=16-12\] \[a=4\] \[\therefore \] Sum of first ten terms is \[{{S}_{10}}=\frac{n}{2}[2a+(n-1)d]\] where \[n=10\] \[=\frac{10}{2}[2\times 4+(10-1)3]\] \[=5[8+27]\] \[=175\]
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