Maths Q5 – JEE-MAIN 2025

Home Maths Sequences and SeriesQuestion #406

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by

\frac{21}{2}

. Then the number of terms which are integers in the A.P. is:

Clue 1 of 3: Define AP properties

Let the AP have $2 n$ terms, first term $a$ , and common difference $d$ . Express the sum of odd and even terms using these variables.

🥷

Ninja Strategy•Calculate Total Terms and Check Integer Condition

First, calculate the total number of terms ( $2 n$ ) to eliminate options. Then, determine if all terms are integers based on $a$ and $d$ to distinguish between the total terms and the count of integer terms.

🧠 Full Solution Path

Step 1: Set up equations from given conditions✦ Active

Let the A.P. have $2 n$ terms, with first term $a$ and common difference $d$ . The sum of the $n$ odd terms is $S_{o d d} = n (a + (n - 1) d) = 24$ . The sum of the $n$ even terms is $S_{e v e n} = n (a + n d) = 30$ . The difference between the last and first term is $a_{2 n} - a_{1} = (2 n - 1) d = \frac{21}{2}$ .

Step 2: Solve for

n

d

, and

a

○ Expand

Subtracting the sum of odd terms from the sum of even terms gives $S_{e v e n} - S_{o d d} = n (a + n d) - n (a + (n - 1) d) = n (d) = 30 - 24 = 6$ . So, $n d = 6$ . Substitute $d = \frac{6}{n}$ into the last term condition: $(2 n - 1) \frac{6}{n} = \frac{21}{2}$ . This simplifies to $12 (2 n - 1) = 21 n \Rightarrow 24 n - 12 = 21 n \Rightarrow 3 n = 12 \Rightarrow n = 4$ . Thus, the total number of terms is $2 n = 8$ . Now find $d = \frac{6}{4} = \frac{3}{2}$ . Substitute $n = 4$ and $d = \frac{3}{2}$ into $S_{o d d} = n (a + (n - 1) d) = 24$ : $4 (a + (4 - 1) \frac{3}{2}) = 24 \Rightarrow a + \frac{9}{2} = 6 \Rightarrow a = \frac{3}{2}$ .

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to avoid errors in solving for $n$ , $d$ , and $a$ .

Step 3: Identify integer terms○ Expand

The terms of the A.P. are given by $a_{k} = a + (k - 1) d$ . Substituting $a = \frac{3}{2}$ and $d = \frac{3}{2}$ , we get $a_{k} = \frac{3}{2} + (k - 1) \frac{3}{2} = \frac{3 k}{2}$ . For $a_{k}$ to be an integer, $3 k$ must be an even number, which implies $k$ must be an even number. Since there are $2 n = 8$ terms, $k$ ranges from $1$ to $8$ . The even values of $k$ are $2, 4, 6, 8$ . Therefore, the integer terms are $a_{2}, a_{4}, a_{6}, a_{8}$ . There are $4$ integer terms in the A.P.

💡 Teacher's Secret Hint

Remember to check the condition for a term to be an integer, not just its value.

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Maths Question 5 – JEE-MAIN 2025

Clue 1 of 3: Define AP properties

🧠 Full Solution Path