Question

If first term is given as \[{{a}_{1}}=4\] and ${(n+1)}^\text{th}$ term as \[{{a}_{n+1}}={{a}_{n}}+4n\] for $n\ge 1$ then the value of \[{{a}_{100}}\] is:
a)19804
b)18904
c)18894
d)19904
e)19894

Answer 1

Hint: In this question we can use the recursion relation between \[{{a}_{n+1}}\] and \[{{a}_{n}}\] repeatedly to find \[{{a}_{100}}\] in terms of \[{{a}_{99}}\], then \[{{a}_{99}}\] in terms of \[{{a}_{99}}\] and so on till we obtain the expression in terms of \[{{a}_{1}}\] whose value is given in the question.

Complete step-by-step solution -
It is given that the value of $a_1$ is 4 i.e. $a_1$=4 and that of $a_{n+1} $ is ${a_n}+{4n}$, where n is any number greater than 1 i.e. \[{{a}_{n+1}}={{a}_{n}}+4n\] for $n\ge 1$. This is known as a recursion relation which enables us to find the value of one term from the value of the other terms
Now, our aim is to obtain \[{{a}_{100}}\] in terms of a1 so that we can find its value from the given value of \[{{a}_{1}}\].
Using the recursion relation (\[{{a}_{n+1}}={{a}_{n}}+4n\] for $n\ge 1$), $a_{100}$ can be written as \[{{a}_{99}}+\text{ }4\times 99\text{ and }{{a}_{99}}\text{ can be written as }{{a}_{98}}+\text{ }4\times 98\] and so on i.e.
\[\begin{align}
  & {{a}_{100}}=\text{ }{{a}_{99}}+\text{ }4\times 99 \\
 & =\left( {{a}_{98}}+\text{ }4\times 98 \right)\text{ }+\text{ }4\times 99\text{ }\ldots \ldots \left( using\text{ }the\text{ }recursion\text{ }relation\text{ }for\text{ }{{a}_{98}} \right) \\
 & =\left( {{a}_{97}}+4\times 97 \right)+\text{ }4\times \left( 98+99 \right) \\
 & ={{a}_{97}}+4\times (97+98+99)..................(1.1) \\
\end{align}\]
And so on till we reach a1 from terms of higher n. Thus, we obtain:
${{a}_{100}}={{a}_{1}}+4\times (1+\ldots +99)$
We notice that in the last term the expression (1+2+3+…+99) is an arithmetic progression with common difference 1. For an arithmetic progression with first and last terms ${{a}_{1}}\text{ and }{{a}_{l}}$ and common difference d, the sum of n terms is given by:
${{s}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+(n-1)d \right)=\dfrac{n}{2}({{a}_{1}}+{{a}_{l}})$
Here, ${{a}_{1}}$=1, ${{a}_{l}}$=99 and n=99. Therefore, we obtain
$1+2+\ldots +99=\dfrac{99}{2}(1+99)=4950$
Using this value in equation (1.1) and using a1=4, we get
${{a}_{100}}={{a}_{1}}+4\times (4950)=4+19800=19804$
Thus, the correct option in the provided question is option a) 19804.

Note: We notice that although in the recursion relation (\[{{a}_{n+1}}={{a}_{n}}+4n\] for $n\ge 1$) the n+1th term is obtained from the nth term by addition of 4n. However, it is not in the form of arithmetic progression because here the added term is different for different n but in arithmetic progression, the same value should be added to each term to get the next term.