question_answer

The 13th term of an AP is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.

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 \[n=10\]

\[=\frac{10}{2}[2\times 4+(10-1)3]\]

\[=5[8+27]\]

\[=175\]