If the first term of an A.P. is 2 and the sum of first five terms is equal to one – fourth the sum of the next five terms, then find the ${{20}^{th}}$ term.

Hint: Consider the first term of the A.P as ‘a’ the common difference as d. Now, find the terms of A.P. using the formula ${{T}_{n}}=a+\left( n-1 \right)d$ where ‘n’ is the ${{n}^{th}}$ term. Take the sum of first five terms, i.e. from ${{T}_{1}}$ to ${{T}_{5}}$ and equate it with $\dfrac{1}{4}$ of the sum of terms from ${{T}_{6}}$ to ${{T}_{10}}$. Form a relation between ‘a’ and ‘d’, substitute the given value a = 2 to find the value of d. Finally, find the value of ${{20}^{th}}$ term by substituting n = 20 in the formula ${{T}_{n}}=a+\left( n-1 \right)d$.

Complete step by step solution:
Here we have been provided with the first term of an A.P. and we are asked to determine its ${{20}^{th}}$ term using the information given. To find the ${{20}^{th}}$ term first we need to find the common difference.
Now, let us consider the first term of the A.P. as ‘a’ and the common difference as ‘d’. We know that the ${{n}^{th}}$ term of an A.P. is given as ${{T}_{n}}=a+\left( n-1 \right)d$ where we can substitute the values of n to get the required terms. So we have,
$\begin{align}
  & {{T}_{1}}=a+\left( 1-1 \right)d=a \\
 & {{T}_{2}}=a+\left( 2-1 \right)d=a+d \\
 & {{T}_{3}}=a+\left( 3-1 \right)d=a+2d \\
\end{align}$
And so on we can find other terms in terms of ‘a’ and ‘d’.
Now, according to the question we have the sum of first five terms equal to one – fourth the sum of the next five terms, that means mathematically,
\[\begin{align}
  & \Rightarrow \left[ {{T}_{1}}+{{T}_{2}}+{{T}_{3}}+{{T}_{4}}+{{T}_{5}} \right]=\dfrac{1}{4}\left[ {{T}_{6}}+{{T}_{7}}+{{T}_{8}}+{{T}_{9}}+{{T}_{10}} \right] \\
 & \Rightarrow \left[ \left( a \right)+\left( a+d \right)+\left( a+2d \right)+\left( a+3d \right)+\left( a+4d \right) \right] \\
 & =\dfrac{1}{4}\left[ \left( a+5d \right)+\left( a+6d \right)+\left( a+7d \right)+\left( a+8d \right)+\left( a+9d \right) \right] \\
 & \Rightarrow \left[ 5a+10d \right]=\dfrac{1}{4}\left[ 5a+35d \right] \\
\end{align}\]
Multiplying both the sides 4 we get,
\[\begin{align}
  & \Rightarrow 20a+40d=5a+35d \\
 & \Rightarrow 20a-5a=35d-40d \\
 & \Rightarrow 15a=-5d \\
\end{align}\]
Dividing both the sides with -5 we get,
\[\Rightarrow d=-3a\]
In the question we have been provided with the value of the first term as 2, so substituting a = 2 in the above relation we get,
\[\begin{align}
  & \Rightarrow d=-3\times 2 \\
 & \Rightarrow d=-6 \\
\end{align}\]
Now, substituting a = 2, d = -6 and n = 20 in the relation ${{T}_{n}}=a+\left( n-1 \right)d$ we get,
$\begin{align}
  & \Rightarrow {{T}_{20}}=2+\left( 20-1 \right)\left( -6 \right) \\
 & \Rightarrow {{T}_{20}}=2+\left( 19 \right)\left( -6 \right) \\
 & \Rightarrow {{T}_{20}}=-112 \\
\end{align}$
Hence ${{20}^{th}}$ is -112.

Note: You must know how to assume successive terms of an A.P. Do not consider them as different variables otherwise you will form many other relations to get the values of different terms. You must also remember the formula of sum of n terms of an A.P. given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. You can also use this sum formula to determine the sum of the first five terms and the next five terms.