Question

The first four terms of an AP are $a,9,3a - b,3a + b$. The ${2011^{th}}$ term the an AP is
1)$2015$
2)$4025$
3)$5030$
4)$8045$

Answer 1

Hint: We are given the first four terms of the AP. So, we can find the common difference of the AP from the first two terms. We know the difference between two consecutive terms of an AP is equal. So, finding the difference between the first and second, second and third, third and fourth terms and then equating these, will give us $a$ and $b$. So, I can find the first term and the common difference of the AP. Then applying the formula for the ${n^{th}}$ term of AP, we can find the ${2011^{th}}$ term of the AP, by using $2011$ in place of n.

Complete step-by-step answer:
The given first four terms of the AP are: $a,9,3a - b,3a + b$.
Therefore, the first term is $a$.
The common difference of the AP can be found out by subtracting the first term from the second term.
So, we get, $d = \left( {9 - a} \right)$
Also, the difference between \[{3^{rd}}\] term and \[{2^{nd}}\] term $ = \left( {3a - b} \right) - 9$
The difference between \[{4^{th}}\] term and ${3^{rd}}$ term $ = \left( {3a + b} \right) - \left( {3a - b} \right)$
Simplifying the expression, we get,
$ = 2b$
We know that the difference between two consecutive terms of an AP is equal.
Therefore, we can say, the difference between \[{2^{nd}}\] and \[{1^{st}}\] term is equal to the difference between \[{3^{rd}}\] and \[{2^{nd}}\] term. So, we get,
$ \Rightarrow 9 - a = 3a - b - 9$
Adding $9$ on both sides of the equation, we get,
$ \Rightarrow 18 - a = 3a - b$
Adding $a$ on both sides of the equation, we get,
$ \Rightarrow 18 = 4a - b$
Changing the sides, we get,
$ \Rightarrow 4a - b = 18 - - - \left( 1 \right)$
Again, we can say, the difference between \[{3^{rd}}\] and \[{2^{nd}}\] term is equal to the difference between \[{4^{th}}\] and \[{3^{rd}}\] term. So, we get,
$ \Rightarrow 3a - b - 9 = 2b$
Subtracting, $2b$ from both sides of the equation, we get,
$ \Rightarrow 3a - 3b - 9 = 0$
Adding $9$ on both sides of the equation, we get,
$ \Rightarrow 3a - 3b = 9$
Dividing both sides by $3$, we get,
$ \Rightarrow a - b = 3 - - - \left( 2 \right)$
Now, subtracting $\left( 2 \right)$ from $\left( 1 \right)$, we get,
$\left( {4a - b} \right) - \left( {a - b} \right) = 18 - 3$
$ \Rightarrow 4a - b - a + b = 15$
Cancelling terms with opposite signs and same magnitude, we get,
$ \Rightarrow 4a - a = 15$
$ \Rightarrow 3a = 15$
Dividing both sides with $3$, we get,
$ \Rightarrow a = 5$
Substituting $a$ in equation $\left( 2 \right)$, we get,
$5 - b = 3$
Subtracting $5$ from both sides, we get,
$ \Rightarrow - b = - 2$
Multiplying both sides with $ - 1$, we get,
$ \Rightarrow b = 2$
Therefore, the first term of the AP $ = a = 5$
The common difference, $d = 9 - a = 9 - 5 = 4$
Therefore, the formula for ${n^{th}}$ term of an AP is, ${t_n} = a + \left( {n - 1} \right)d$
So, to find the ${2011^{th}}$ term, replacing $n = 2011$ and substituting the values of $a$ and $d$, we get,
${t_{2011}} = 5 + \left( {2011 - 1} \right)4$
$ \Rightarrow {t_{2011}} = 5 + \left( {2010} \right)4$
$ \Rightarrow {t_{2011}} = 5 + 8040$
$ \Rightarrow {t_{2011}} = 8045$
Therefore, the ${2011^{th}}$ term of the AP is $8045$, the correct option is 4.
So, the correct answer is “Option 4”.

Note: In some questions, we are just told that $4$ numbers are in AP. So, in such cases, it is better for the sake of calculation and simplification that we assume the four numbers of AP to be $a - 3d,a - d,a + d,a + 3d$. Similarly, if three numbers are in AP, we assume the numbers to be $a - d,a,a + d$. Taking these terms helps us with the calculations as they become easier.