Question

How many terms of A.P 9, 17, 25, …. Must be taken to give the sum of 450?

Answer 1

Hint: We solve this problem by using the sum of terms of an A.P that is the sum of \['n'\] terms of A.P having first term \['a'\] and the common difference \['d'\] is given as \[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
By using the above formula we calculate the number of terms required to get a sum of 450 for given A.P

Complete step-by-step solution
We are given that the A.P as
9, 17, 25, …..
Here we can see that the first term of A.P that is
\[\Rightarrow a=9\]
We know that the common difference of A.P is given by subtracting first term from second term that is
\[\Rightarrow d=17-9=8\]
We are given that the sum of \['n'\] terms of this A.P as 450
We know that the sum of \['n'\] terms of A.P having first term \['a'\] and the common difference \['d'\] is given as
\[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
By using the above formula to given A.P we get
\[\begin{align}
  & \Rightarrow 450=\dfrac{n}{2}\left( 2\times 9+\left( n-1 \right)8 \right) \\
 & \Rightarrow 450=\dfrac{2n}{2}\left( 9+4n-4 \right) \\
 & \Rightarrow 4{{n}^{2}}+5n-450=0 \\
\end{align}\]
Now let us solve the above quadratic equation using the factorisation method that is let us rewire the middle term in such a way that we can get the factors out that is
\[\Rightarrow \left( 4{{n}^{2}}+45n \right)+\left( -40n-450 \right)=0\]
Now, by taking out the common terms out from both brackets we get
\[\begin{align}
  & \Rightarrow n\left( 4n+45 \right)-10\left( 4n+45 \right)=0 \\
 & \Rightarrow \left( 4n+45 \right)\left( n-10 \right)=0 \\
\end{align}\]
We know that if \[a\times b=0\] then either \['a'\] or \['b'\] will be zero.
By using the above condition we get the first term as
\[\Rightarrow n=\dfrac{-45}{4}\]
Similarly, by taking the second term we get
\[\Rightarrow n=10\]
We know that the value of \['n'\] will never be negative and fraction because it is a number of terms.
Therefore the value of \['n'\] will be 10
So, we can conclude that 10 terms must be taken from a given A.P to get the sum as 450.

Note: We can solve the quadratic equation in other methods also.
We have the equation as
\[\Rightarrow 4{{n}^{2}}+5n-450=0\]
We know that the roots of equation of form \[a{{x}^{2}}+bx+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By using the above formula we get
\[\begin{align}
  & \Rightarrow n=\dfrac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 4 \right)\left( -450 \right)}}{2\times 4} \\
 & \Rightarrow n=\dfrac{-5\pm \sqrt{7225}}{8} \\
 & \Rightarrow n=\dfrac{-5\pm 85}{8} \\
\end{align}\]
Here we have two roots for \['n'\] considering positive sign once and negative sign once.
We know that the number of terms will never be negative so, the possible value for \['n'\] will be taken from positive sign that is
\[\begin{align}
  & \Rightarrow n=\dfrac{-5+85}{8} \\
 & \Rightarrow n=10 \\
\end{align}\]
Therefore the value of \['n'\] will be 10
So, we can conclude that 10 terms must be taken from a given A.P to get the sum as 450.