Question

The maximum number of terms common to the arithmetic progression 3 ,7 ,11 ,15 ,19,23 ,...............403. and 5 ,11, 17 ,23 ,29, ..............505 is ?

Answer 1

Thank you for the question.

First of all we will take the first sequence. 3,7,11,15,19,24,...403.
We will find the common difference between the terms. Between 3 and 7 , 7 and 11, and so on, the difference is 4. 
So we will denote d=4

Next we will identify the first and last term and denote it by A and L respectively.
A=3 and L= 403.

Lastly, we will substitute them in the following formula to get out answer:

n= (L - A)/d +1
n= (403 - 3)/4 +1
n= 400/4 +1
n= 100 + 1
n= 101

We will repeat the procedure for sequence 2. 
5,11,17,23,29...505

d=6
A=5
L=505

n=(L - A)/d +1
n=(505 - 5)/6 +1
n=500/6 + 1
n= 83.3 + 1
n= 84.3

Answer 2

L₁: 3, 7, 11, 15, 19 , 23,........403 and L₂: 5, 11, 17, 23, 29, ........505
Let nth term of L₁ is equal to mth term of L₂
For L₁,
a = 3, d = 4
Tn = a + (n -1)d
Tn = 3 + (n -1)4 = 4n -1

For L₂,
a = 5 , d = 11
Tm = a + (m -1)d
Tm = 5 + (m -1)6 = 6m -1

now, Tn = Tm
4n -1 = 6m -1
4n = 6m ⇒n/3= m/2 = K (let)
n = 3K , m = 2K
now, number of terms in L₁ : Tn = a + (n-1)d
403 = 3 + (n-1)4 ⇒n = 101
number of terms in L₂ : Tm = a + (m-1)d
505 = 5 + (m-1)6 ⇒m = 250/3 + 1= 84.33

Now, for last common number condition will be n = 3k ≤ 101 and m = 2k ≤ 84.33
3k ≤101 ⇒K ≤ 33.33 and 2k≤ 84.33 ⇒K ≤ 42.16
Hence, K ≤ 33.33 so, maximum value of K = 33 in integer

Hence, maximum number of terms common to the arithmetic progression is 33