JEE Mathematics 2023 Question with Solution

Given: $a_5 = 2a_1$ and $a_1 = 18$.

Find: The value of

$$12 \left( \frac{1}{\sqrt{a_0} + \sqrt{a_1}} + \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \cdots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}} \right)$$

From the solution:

$$a_n = a_1 + (n-1)d$$

Using $a_5 = a_1 + 4d$ and $a_5 = 36$,

$$36 = 18 + 4d$$

so

$$d = \frac{18}{4} = 4.5$$

Also,

$$a_{18} = a_1 + 17d = 18 + 17 \times 4.5 = 94.5$$

The provided solution then states that the series simplifies and gives the final result as

$$12 \times 9 = 108$$

Therefore, the required value is $108$.

Given: $a_5 = 2a_1$ and $a_1 = 18$.

Find: The required numerical value.

1. Since $a_1 = 18$, we get

$$a_5 = 2 \times 18 = 36$$

2. For an arithmetic progression,

$$a_n = a_1 + (n-1)d$$

3. Using $a_5 = a_1 + 4d$,

$$36 = 18 + 4d$$

Hence,

$$18 = 4d$$

and therefore,

$$d = \frac{18}{4} = 4.5$$

4. Now,

$$a_{18} = a_1 + 17d$$

Substituting the values,

$$a_{18} = 18 + 17 \times 4.5 = 94.5$$

5. The solution next states that the summation is simplified term by term and concludes that the total inside the bracket is $9$, so the whole expression becomes

$$12 \times 9 = 108$$

Therefore, the answer is $108$.

Note: The intermediate simplification of the radical sum is not fully shown in the solution, so only the displayed working has been extracted.