Answer:
We know that \[{{T}_{n}}=a+(n-1)d\] Given, \[{{T}_{4}}=a+(4-1)d=0\] \[a+3d=0\] \[a=-3d\] \[{{T}_{25}}=a+(25-1)d\] \[=a+24d=(-3d)+24d\] \[=24d\] And, \[{{T}_{11}}=a+(11-1)d\] \[=a+10d\] Then, \[3{{T}_{11}}=3(a+10d)\] \[=3a+30d\] \[=3(-3d)+30d(a=-3d)\] \[=30d-9d=21d={{T}_{25}}\] \[\therefore \] \[3{{T}_{11}}={{T}_{25}}\] Hence Proved.
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