Question

The number of terms common of two A.P’ s 5, 9, 13, ………, 2013 and 2, 9, 16, ……., 3516 is
A) 71
B) 70
C) 72
D) 68

Answer 1

Hint:
We can find the common difference and \[{1^{st}}\] term of both the AP. Then we can find a new AP with the \[{1^{st}}\] term as the \[{1^{st}}\] term common for both the AP and the new common difference as the product of the common difference of the 2 APs. Then we can take the last term of the smaller AP and take it as the nth term. Then we can approximate the value of n using the formula for ${n^{th}}$ term. Then the integer lesser than or equal to the approximate value will be the required number of terms.

Complete step by step solution:
We have the \[{1^{st}}\] AP as 5, 9, 13, …., 2013
Here the \[{1^{st}}\] term is $a = 5$and the common difference is given by the difference between two consecutive terms.
$ \Rightarrow {d_1} = 9 - 5$
So, we have
$ \Rightarrow {d_1} = 4$
Now consider the ${2^{nd}}$ AP, 2, 9, 16, …, 3516
Here the \[{1^{st}}\] term is $a = 2$ and the common difference is given by the difference between two consecutive terms.
$ \Rightarrow {d_2} = 9 - 2$
So, we have
$ \Rightarrow {d_2} = 7$
Now we can consider a new AP with the terms common in both the APs.
The \[{1^{st}}\] term of the new AP is the \[{1^{st}}\] occurring term which is common in both the APs. From the APs, the 9 is the \[{1^{st}}\] term of the new AP.
$ \Rightarrow a = 9$
Now the common difference of the new AP is given by the product of the common difference of the 2 Aps.
$ \Rightarrow d = {d_1} \times {d_2}$
On substituting the values, we get
$ \Rightarrow d = 4 \times 7$
So, we have
$ \Rightarrow d = 28$
As the new AP has the terms that are common to both the terms, it will have the terms only in the smaller AP, that is the \[{1^{st}}\] AP.
So, let us take the last term of the new AP as the last term of the \[{1^{st}}\] AP. Let the number of terms in new AP be m.
$ \Rightarrow {a_m} = 2013$
 We know that the \[{n^{th}}\] term of the AP is given by ${a_m} = a + \left( {m - 1} \right)d$ . On substituting the values, we get
$ \Rightarrow 2013 = 9 + \left( {m - 1} \right)28$
On rearranging, we get
$ \Rightarrow 2013 - 9 = \left( {m - 1} \right)28$
On dividing the equation by 28, we get
$ \Rightarrow \left( {m - 1} \right) = \dfrac{{2004}}{{28}}$
So, we have
$ \Rightarrow \left( {m - 1} \right) = 71.57$
On adding 1 on both sides, we get
$ \Rightarrow m = 71.57 + 1$
On addition, we get
$ \Rightarrow m = 72.57$
We know that the number of terms in an AP cannot be a decimal. So, 2013 is not a term of the AP. So, the number of terms in the AP will be the integer less than $72.57$. Therefore, the number of terms in the new AP is 72.
Thus, the number of terms common of two A.P.’s 5, 9, 13, …., 2013 and 2, 9, 16, …, 3516 is 72

So, the correct answer is option C.

Note:
We must not conclude that there is only one common term as there is only one common term in the representation of the AP in the question. We must keep in mind that the number of terms is always a natural number. We must take the greatest natural number that is less than or equal to the value of m we found as the number of terms in the new AP. We must not take the natural number greater than m as we equated the last term as the highest possible number that can be in the AP.