JEE Main 2025 — 29 January, Evening Shift. Previous Year Question.
Problem
Let $a_1,a_2,\ldots,a_{2024}$ be an Arithmetic Progression such that
$$a_1 + \big(a_5+a_{10}+a_{15}+\cdots+a_{2020}\big) + a_{2024} = 2233$$
Then $a_1+a_2+a_3+\cdots+a_{2024}$ is equal to ____.
Key insight. The terms $a_5, a_{10}, \ldots, a_{2020}$ are themselves evenly spaced — they form their own smaller AP, with common difference $5d$ instead of $d$. Summing that smaller AP with the standard formula, then adding $a_1$ and $a_{2024}$, expresses the given condition purely in terms of $2a_1+2023d$ — which is exactly the quantity needed to find the full sum of 2024 terms.
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Approach
Let the AP have first term $a$ and common difference $d$. Recognise that $a_5, a_{10}, \ldots, a_{2020}$ is itself an arithmetic progression, count how many terms it has, and sum it using the standard AP sum formula. Combine this with $a_1$ and $a_{2024}$ to form one equation in $2a+2023d$, then use that same quantity to compute the full sum directly — without ever solving for $a$ and $d$ separately.
Solution
Step 1 — Identify the smaller AP
The subscripts $5,10,15,\ldots,2020$ increase by $5$ each time, so $a_5,a_{10},\ldots,a_{2020}$ is itself an AP with first term $a_5=a+4d$ and common difference $5d$.
Step 2 — Count the number of terms
The number of terms from $5$ to $2020$ in steps of $5$ is:
$$\frac{2020-5}{5}+1 = 404$$
Step 3 — Sum the smaller AP
$$\sum = \frac{404}{2}\Big[2(a+4d) + (404-1)(5d)\Big] = 202\big[2a+8d+2015d\big] = 202(2a+2023d)$$
Step 4 — Add a1 and a2024, and use the given condition
$$a_1 + \sum + a_{2024} = a + 202(2a+2023d) + (a+2023d) = 2233$$
$$2a+2023d + 202(2a+2023d) = 2233$$
$$203(2a+2023d) = 2233$$
Step 5 — Solve for the key quantity
$$2a+2023d = \frac{2233}{203} = 11$$
Step 6 — Find the sum of all 2024 terms
The sum of the full AP uses exactly this quantity:
$$S_{2024} = \frac{2024}{2}\big[2a+(2024-1)d\big] = 1012(2a+2023d) = 1012 \times 11$$
$$S_{2024} = 11132$$
Answer
$$11132$$
Common mistakes
- Trying to solve for $a$ and $d$ individually. The problem only gives one equation, so $a$ and $d$ can’t be found separately — but the combination $2a+2023d$ can be found, and it happens to be exactly what’s needed for the final sum. Chasing individual values wastes time and may seem impossible.
- Miscounting the number of terms in $a_5,a_{10},\ldots,a_{2020}$. It’s tempting to divide $2020$ by $5$ directly to get $404$, but the correct count is $\frac{2020-5}{5}+1$ — these happen to agree here, but only because the sequence starts at $5$ rather than $0$; always verify with the “$(last-first)/step + 1$” formula.
Practise next
- For an AP $a_1,\ldots,a_{1000}$, if $a_1+a_5+a_{996}+a_{1000}=1500$, find $a_1+a_2+\cdots+a_{1000}$ using the same sub-AP summation method.
Show answer
$375000$. The indices pair up symmetrically: $1+1000=5+996=1001$, and in an AP any two terms whose indices sum to the same value have the same total.
So $a_1+a_{1000}=a_5+a_{996}=2a_1+999d$, and the given sum is $2(2a_1+999d)=1500$, giving $2a_1+999d=750$.
That is exactly the bracket in $S_{1000}=\dfrac{1000}{2}(2a_1+999d)=500\times750=375000$. Neither $a_1$ nor $d$ is ever needed on its own.
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