Question

Let AP (a, d) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d>0. If \[AP\left( 1,3 \right)\cap AP\left( 2,5 \right)\cap AP\left( 3,7 \right)=AP\left( a,d \right)\] then a+d equals

Answer 1

Hint: To solve this question, we will first calculate ${{m}^{th}},{{k}^{th}}\text{ and }{{\text{t}}^{\text{th}}}$ terms of AP (1, 3), AP (2, 5) and AP (a, d) respectively. This is done by using formula of ${{n}^{th}}$ term of an AP which is given as:
\[an=a+\left( n-1 \right)d\]
Where, a = first term, an = ${{n}^{th}}$ term, d = common difference and n = number of terms.
After doing so, we will try to find a relation between m, k and t and calculate one variable in terms of the other two. Again, we will use the ${{n}^{th}}$ term formula of AP in the AP (2, 5) whose variable is assumed to get a and d is 1cm LCM of common difference of all three given AP.

Complete step-by-step answer:
We are given an infinite arithmetic progression AP (a, d) with first term a and common difference $d>0$
We have formula for ${{n}^{th}}$ term of a arithmetic progression AP as:
\[an=a+\left( n-1 \right)d\]
Where, a = first term, an = ${{n}^{th}}$ term, d = common difference and n = number of terms.
We are given that \[AP\left( 1,3 \right)\cap AP\left( 2,5 \right)\cap AP\left( 3,7 \right)=AP\left( a,d \right)\]
So, let us first calculate ${{m}^{th}}$ term of AP (1, 3)
AP (1, 3) has a = 1, d = 3
Then, ${{m}^{th}}$ term using above formula is
\[\begin{align}
  & {{m}^{th}}\text{ term}=1+\left( n-1 \right)3 \\
 & {a}_{m}=1+\left( n-1 \right)3\Rightarrow 3n-2 \\
 & {a}_{m}=3n-2\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
Where am is ${{m}^{th}}$ term.
Again similarly we will calculate ${{k}^{th}}$ term of AP (2, 5)
Let ${{a}_{k}}$ be ${{k}^{th}}$ term of AP (3, 7)
Then, using above formula we get:
\[\begin{align}
  & {{a}_{k}}=2+\left( k-1 \right)5 \\
 & {{a}_{k}}=2+5k-5 \\
 & {{a}_{k}}=5k-3\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
Similarly, we will calculate ${{t}^{th}}$ term of AP (3, 7)
Let ${{a}_{t}}$ be ${{t}^{th}}$ term of AP (3, 7)
Then, using above formula we get:
\[\begin{align}
  & {{a}_{t}}=3+\left( t-1 \right)7 \\
 & {{a}_{t}}=3+7t-7 \\
 & {{a}_{t}}=7t-4\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)} \\
\end{align}\]
Now, all the three terms obtained in equation (i), (ii) and (iii) are equal by the fact that, there intersection is a common AP(a, d).
\[\Rightarrow 3n-2=5k-3=7t-4\]
Consider the first two terms then
\[\begin{align}
  & 3n-2=5k-3 \\
 & 3n=5k-3+2 \\
 & 3n=5k-1 \\
\end{align}\]
Dividing by 3,
\[n=\dfrac{5k-1}{3}\]
Now, because n is the number of term, so it cannot be a fraction, it needs to be an integer.
Then all possible values of k such that n is an integer is
\[k=2,5,11...\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)}\]
Again consider second and third part of
\[\begin{align}
  & 5k-3=7t-4 \\
 & 5k=7t+3-4 \\
 & 5k+1=7t \\
 & 7t=5k+1 \\
 & t=\dfrac{5k+1}{7} \\
\end{align}\]
Again as t is number of terms, so all possible values of R such that t is an integer is
\[k=4,11....\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (v)}\]
First common values of R as obtained in equation (iv) and (v) is k = 11.
Then, calculating the first term a by using formula of ${{n}^{th}}$ in AP(2, 5)
Then the value of a (first term) is:
\[\begin{align}
  & a=2+\left( k-1 \right)5 \\
 & a=2+\left( 11-1 \right)5 \\
 & a=2+10\times 5 \\
 & a=52 \\
\end{align}\]
So, first term a = 52.
And the common difference d is nothing but the LCM of common difference of AP (1, 3), AP (2, 5) and AP (3, 7)
\[\begin{align}
  & d=LCM\left( 3,5,7 \right) \\
 & 3\left| \!{\underline {\,
  3,5,7 \,}} \right. \\
 & 5\left| \!{\underline {\,
  1,5,7 \,}} \right. \\
 & 7\left| \!{\underline {\,
  1,1,7 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1,1,1 \,}} \right. \\
\end{align}\]
So, LCM of $\left( 3,5,7 \right)=3\times 5\times 7=105$
d = 105
Then, value of a + d = 105 + 52 = 157
Therefore, the value of a + d = 157

Note: The possibility of confusion in this question can be at the point that, why AP (2, 5) is only considered to get value of first term a.
This is so because we have calculated the value of t and m both in terms of k. So, first term a is also given by using k series or AP (2, 5)
As we had calculated m in terms of k and t then, we had used AP (1, 3) to calculate a first term.
Answer would anyway be the same, so this confusion is cleared.