• # question_answer 9) The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

 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.