Question

In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that \[{{20}^{th}}\] term is -112.

Answer 1

Hint: The formula for writing \[{{n}^{th}}\] term of an arithmetic progression is \[{{n}^{th}}\ term=a+(n-1)d\] (where ‘a’ is the first term and‘d’ is the common difference of the arithmetic progression). The formula for writing the sum of first n terms of an arithmetic progression is $S_n$\[=\dfrac{n}{2}\left( 2a+(n-1)d \right)\]

Complete step-by-step answer:

As mentioned in the question, it is given that the first term of the arithmetic progression is 2, therefore a is equal to 2. It is also given in the that the sum of first five terms is one-fourth of the sum of the next five terms that is
\[SU{{M}_{1-5}}=\dfrac{1}{4}SU{{M}_{6-10}}\]
Using the sum formula as mentioned in the hint, we get
\[\begin{align}
  & SU{{M}_{1-5}}=\dfrac{5}{2}\left( 2\times 2+(5-1)d \right) \\
 & \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{5}{2}\left( 2\times 2+4d \right) \\
 & \ \ \ \ \ \ \ \ \ \ \ \ =5(2+2d) \\
 & \ \ \ \ \ \ \ \ \ \ \ \ =10+10d\ \ \ \ \ ...(a) \\
\end{align}\]
(Here,‘d’ is the common difference of the arithmetic progression)
For the \[SU{{M}_{6-10}}\] , we will have to write it as
\[SU{{M}_{6-10}}=SU{{M}_{1-10}}-SU{{M}_{1-5}}\ \ \ \ \ ...(b)\]
Hence, the given condition can be written using equation (a) and (b) as\[\begin{align}
  & SU{{M}_{1-5}}=\dfrac{1}{4}SU{{M}_{6-10}} \\
 & 10+10d=\dfrac{1}{4}\left( SU{{M}_{1-10}}-SU{{M}_{1-5}} \right) \\
 & 10+10d=\dfrac{1}{4}\left[ \left( \dfrac{10}{2}(2\times 2+(10-1)d) \right)-\left( \dfrac{5}{2}(2\times 2+(5-1)d) \right) \right] \\
 & 10+10d=\dfrac{1}{4}\left[ \left( 5(4+9d) \right)-\left( 5(2+2d) \right) \right] \\
 & 10+10d=\dfrac{1}{4}\left[ \left( 20+45d) \right)-\left( 10+10d) \right) \right] \\
 & 10+10d=\dfrac{1}{4}\left[ \left( 10+35d \right) \right] \\
 & 40+40d=10+35d \\
 & 5d=-30 \\
 & d=-6 \\
\end{align}\]
Now, using the formula for the \[{{n}^{th}}\] term for finding the \[{{20}^{th}}\] term is
\[\begin{align}
  & {{20}^{th}}\ term=2+(20-1)d \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ =2+19\times (-6) \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ =-112 \\
\end{align}\]
Hence, the 20th term is -112.

Note: The students can make an error in writing the sum and \[{{n}^{th}}\] term if they might confuse in finding the common difference that is ‘d’ as the value of ‘d’ would be found only on proceeding with the question by taking the common difference as an unknown variable.