question_answer

The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

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.