JEE Mathematics 2023 Question with Solution

Given: Arithmetic Progression: $3, 8, 13, \ldots, 373$

Find: The sum of all terms which are not divisible by $3$.

First, find the total number of terms:

$$T_n = a + (n - 1)d$$

Substitute $T_n = 373$, $a = 3$, $d = 5$:

$$373 = 3 + (n - 1)5$$ $$370 = 5(n - 1)$$ $$n - 1 = \frac{370}{5} = 74$$ $$n = 75$$

Now find the sum of the full arithmetic progression:

$$\text{Sum} = \frac{n}{2}[a + l]$$

Substitute $n = 75$, $a = 3$, $l = 373$:

$$\text{Sum} = \frac{75}{2}[3 + 373] = \frac{75}{2}(376) = 75 \cdot 188 = 14100$$

Next, find the terms divisible by $3$. These are $3, 18, 33, \ldots, 363$.

Find their number:

$$363 = 3 + (k - 1)15$$ $$360 = (k - 1)15$$ $$k - 1 = \frac{360}{15} = 24$$ $$k = 25$$

Now find the sum of these terms:

$$\text{Sum} = \frac{k}{2}[a + l]$$

Substitute $k = 25$, $a = 3$, $l = 363$:

$$\text{Sum} = \frac{25}{2}[3 + 363] = \frac{25}{2}(366) = 25 \cdot 183 = 4575$$

Therefore, the required sum is:

$$\text{Required Sum} = 14100 - 4575 = 9525$$

Therefore, the required answer is $9525$.

Given: The arithmetic progression $3, 8, 13, \ldots, 373$

Find: Sum of terms not divisible by $3$.

The progression has first term $a = 3$ and common difference $d = 5$.

Using the last-term formula:

$$373 = 3 + (n - 1)5$$ $$373 - 3 = 5(n - 1)$$ $$370 = 5(n - 1)$$ $$n - 1 = 74$$ $$n = 75$$

So the sum of all $75$ terms is:

$$S = \frac{75}{2}(3 + 373)$$ $$S = \frac{75}{2}(376)$$ $$S = 14100$$

Now identify terms divisible by $3$. Since the AP increases by $5$, every third term is divisible by $3$, giving the sequence:

$$3, 18, 33, \ldots, 363$$

This is also an AP with first term $3$ and common difference $15$.

Its number of terms is found from:

$$363 = 3 + (k - 1)15$$ $$360 = 15(k - 1)$$ $$k - 1 = 24$$ $$k = 25$$

Its sum is:

$$S_1 = \frac{25}{2}(3 + 363)$$ $$S_1 = \frac{25}{2}(366)$$ $$S_1 = 4575$$

Subtract these divisible terms from the total sum:

$$14100 - 4575 = 9525$$

So, the sum of all terms not divisible by $3$ is $9525$.