Question

Insert the missing number in the given series:
0, 4, 18, 48, ?, 180
(a) 58
(b) 68
(c) 84
(d) 100

Answer 1

We solve this problem first by finding the general representation of each term.
That is we find the representation of ${{n}^{th}}$ in terms of $'n'$ to find the required term. If the ${{n}^{th}}$ term is denoted as ${{T}_{n}}$ then we can find the value of ${{5}^{th}}$ term by substituting $n=5$ which gives the value of ${{5}^{th}}$ term as ${{T}_{5}}$ which is the required answer.

Complete step by step answer:
We are given that series as
0, 4, 18, 48, ?, 180
Let us assume that the value of ${{5}^{th}}$ term as $'x'$
Now, let us try to find the general term that is we find the representation of ${{n}^{th}}$ term in terms of $'n'$
Now, let us take the first term that is 0 as follows
${{1}^{st}}$ term is given as $\left( 1-1 \right)\times {{1}^{2}}$
Now, let us take the second term that is 4 as follows
${{2}^{nd}}$ term is given as $\left( 2-1 \right)\times {{2}^{2}}$
Now, let us take the third term that is 18 as follows
${{3}^{rd}}$ term is given as $\left( 3-1 \right)\times {{3}^{2}}$
Now, let us take the fourth term that is 48 as follows
${{4}^{th}}$ term is given as $\left( 4-1 \right)\times {{4}^{2}}$
Now, let us assume that the ${{n}^{th}}$ term is given as ${{T}_{n}}$
Now, we can take the ${{n}^{th}}$ term from the first four terms as

& \Rightarrow {{T}_{n}}=\left( n-1 \right)\times {{n}^{2}} \\\ & \Rightarrow {{T}_{n}}={{n}^{3}}-{{n}^{2}} \\\ \end{aligned}$$ Now, let us find the $${{5}^{th}}$$ term by substituting $$n=5$$ in above equation we get $$\begin{aligned} & \Rightarrow {{T}_{5}}={{5}^{3}}-{{5}^{2}} \\\ & \Rightarrow {{T}_{5}}=125-25 \\\ & \Rightarrow {{T}_{5}}=100 \\\ \end{aligned}$$ Therefore we can conclude that the missing term is 100 So, we can have the complete series as 0, 4, 18, 48, 100, 180 **So, the correct answer is “Option d”.** **Note:** Students may misunderstand the given series with arithmetic – geometric progression (AGP). We are given that the series as 0, 4, 18, 48, … The given series can be modified as $$\left( 0\times {{1}^{2}} \right),\left( 1\times {{2}^{2}} \right),\left( 2\times {{3}^{2}} \right),\left( 3\times {{4}^{2}} \right),.....$$ Students may take this series as AGP but it is not AGP. The general representation of AGP is given as $$\left( a \right),\left( a+d \right){{r}^{2}},\left( a+2d \right){{r}^{3}},.....$$ Here in the given series the first terms in the product are in AP but the second terms in the product are not in GP. So, we can say that the given series is not an AGP.