JEE Mathematics 2023 Question with Solution

Given: The three arithmetic progressions are

$$\text{AP}_1 : 3, 7, 11, 15, \ldots, 399 \quad (d_1 = 4)$$ $$\text{AP}_2 : 2, 5, 8, 11, \ldots, 359 \quad (d_2 = 3)$$ $$\text{AP}_3 : 2, 7, 12, 17, \ldots, 197 \quad (d_3 = 5)$$

Find: The sum of the common terms of these three APs.

To find common terms, use the least common multiple of the common differences:

$$\text{LCM}(4,3,5)=60$$

The common terms are:

$$47, 107, 167$$

Their sum is:

$$47+107+167=321$$

Therefore, the required sum is $321$.

Given: Three APs with common differences $4$, $3$, and $5$.

Find: The sum of terms that are present in all three progressions.

A term common to all three APs must recur with step size equal to the LCM of the three differences:

$$\text{LCM}(4,3,5)=60$$

So once one common term is found, the next common terms differ by $60$.

From the listed terms, the common terms identified are:

$$47, 107, 167$$

All of them lie within the given ending terms $399$, $359$, and $197$ respectively.

Now add them:

$$\begin{aligned} 47+107 &= 154 \\ 154+167 &= 321 \end{aligned}$$

Hence, the sum of the common terms is $321$.