question_answer

The 16th term of an AP is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.

Answer:

Given that 16th term of an A.P is five time its 3rd term.

i.e.,                   \[a+(16-1)d=5[a+(3-1)d]\]

\[\Rightarrow a+15d=5[a+2d]\]

\[\Rightarrow a+15d=5a+10d\]

\[\Rightarrow 4a-5d=0\]                                       ?(i)

Also given that,

\[{{a}_{10}}=41\]

\[\Rightarrow a+(10-1)d=41\]

\[\Rightarrow a+9d=41\]                                     ...(ii)

On multiplying equation (ii) by 4, we get

\[4a+36d=164\]                                     ...(iii)

Subtracting equation (iii) from (i), we get

\[_{\begin{smallmatrix}  4a\,\,+\,\,36\,\,d\,\,=\,\,164 \\  -\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-  \\  \overline{\,\,\,\,\,\,\,\,\,\,-\,41\,d\,\,=\,\,-164} \end{smallmatrix}}^{4a\,\,-\,\,\,\,5d\,\,\,\,\,=\,\,0}\]

\[d=4\]

On putting the value of d in eq. (i), we get

\[4a-5\times 4=0\]

\[4a=20\]

\[a=5\]

Now,                 \[{{S}_{15}}=\frac{15}{2}[2a+(15-1)d]\]

\[{{S}_{15}}=\frac{15}{2}(2\times 5+14\times 4)\]

\[=\frac{15}{2}2(5+14\times 2)\]

\[=15(5+28)\]

\[=15\times 33\]

\[{{S}_{15}}=495\]

Hence, sum of first fifteen terms is \[495\].