Question

The two series of Arithmetic Progression are $2,4,6,......,100$ and $3,6,9,......,99$. How many common terms are there between the two A.P.

Answer 1

We solve this question by first considering the first A.P and find the common difference and then the number of terms in the progression by equating the last term to the formula of ${{n}^{th}}$ term of an A.P, ${{a}_{n}}=a+\left( n-1 \right)d$. Then we find the general term of this A.P. Then we consider the second A.P and repeat the same process as above and find the general term form for it. Then we equate the both general terms obtained and find the condition for which they are true. The obtained condition is the number of terms common to both A.P.

Complete step by step answer:
We are given two series of Arithmetic Progressions $2,4,6,......,100$ and $3,6,9,......,99$.
Now let us consider the first Arithmetic Progression, $2,4,6,......,100$.
In it we can see that the first term is 2.
The common difference of the A.P is

& \Rightarrow \text{Second Term}-\text{First Term} \\\ & \Rightarrow 4-2 \\\ & \Rightarrow 2 \\\ \end{aligned}$$ So, we have that the first term of the A.P is 2 and the common difference of the A.P is 2. Now let us consider the formula for the ${{n}^{th}}$ term of an A.P with first term $a$ and common difference $d$. ${{a}_{n}}=a+\left( n-1 \right)d$ Using this formula, we can find the place of last term of the A.P, that is for 100. $\begin{aligned} & \Rightarrow 2+\left( n-1 \right)2=100 \\\ & \Rightarrow \left( n-1 \right)2=98 \\\ & \Rightarrow n-1=49 \\\ & \Rightarrow n=50 \\\ \end{aligned}$ So, the total number of terms in the first progression is 50. Using the formula, we can say that any term of this A.P is of the form, $\begin{aligned} & \Rightarrow 2+\left( n-1 \right)2\,\ \ \ where\ \ n\le 50 \\\ & \Rightarrow 2+2k\,\ \ \ where\ \ k\le 49.............\left( 1 \right) \\\ \end{aligned}$ Now let us consider the second Arithmetic Progression, $3,6,9,......,99$. In it we can see that the first term is 3. The common difference of the A.P is $$\begin{aligned} & \Rightarrow \text{Second Term}-\text{First Term} \\\ & \Rightarrow 6-3 \\\ & \Rightarrow 3 \\\ \end{aligned}$$ So, we have that the first term of the A.P is 3 and the common difference of the A.P is 3. Now let us consider the formula for the ${{n}^{th}}$ term of an A.P with first term $a$ and common difference $d$. ${{a}_{n}}=a+\left( n-1 \right)d$ Using this formula, we can find the place of last term of the A.P, that is for 99. $\begin{aligned} & \Rightarrow 3+\left( n-1 \right)3=99 \\\ & \Rightarrow \left( n-1 \right)3=96 \\\ & \Rightarrow n-1=32 \\\ & \Rightarrow n=33 \\\ \end{aligned}$ So, total number of terms in the second progression is 33. Using the formula, we can say that any term of this A.P is of the form, $\begin{aligned} & \Rightarrow 3+\left( n-1 \right)3\,\ \ \ where\ \ n\le 33 \\\ & \Rightarrow 3+3p\,\ \ \ where\ \ p\le 32.............\left( 2 \right) \\\ \end{aligned}$ Now let us find the number of terms that are common to both the A.P. If they are common then they must satisfy the equations (1) and (2). So, let us equate them. So, the common terms are the values that satisfying the condition, $\begin{aligned} & \Rightarrow 2+2k=3+3p\,\ \ \ where\ \ p\le 32\,,k\le 49 \\\ & \Rightarrow 2\left( 1+k \right)=3\left( 1+p \right)\,\ \ \ where\ \ p\le 32\,,k\le 49 \\\ & \Rightarrow 1+k=\dfrac{3}{2}\left( 1+p \right)\,\ \ \ where\ \ p\le 32\,,k\le 49 \\\ \end{aligned}$ As $k$ is an integer, $\dfrac{3}{2}\left( 1+p \right)$ must be an integer, which is possible if and only if $\left( 1+p \right)$ is divisible by 2. So, the required terms are the number of times $\left( 1+p \right)$ is even when $p\le 32$, that is when $p$ is odd when $p\le 32$. When $p\le 32$, $p$ is odd 16 times. So, the number of terms common for both the series are 16. **So, the correct answer is “16”.** **Note:** We can also solve this question in a simple alternate process. The two A.P we are given are $2,4,6,......,100$ and $3,6,9,......,99$. Now we can see that the first progression is multiples of 2 less than are equal to 100. Similarly, the second progression is multiples of 3 that are less than 100. So, the terms that are common to both of them are the terms that are less than 100 and divisible by both 3 and 2.