Suppose that the number of terms in an A.P. is 2k, k ∈ N. If the sum of all odd terms of the A.P. is

Q: Why subtract the sums instead of solving for $a_1$ and $d$ separately?

In JEE Advanced, time is your currency. Solving for $a_1$ and $d$ creates a system of two variables that is unnecessary. By subtracting the sums, we isolate the product $kd$ directly, which is the 'soul' of the problem. It is a shortcut that demonstrates mathematical maturity.

Q: Is it always true that $S_{even} - S_{odd} = kd$?

Yes, for an A.P. with $2k$ terms. Since each pair $(a_{2n} - a_{2n-1})$ equals $d$, and there are $k$ such pairs, the result is always $k \cdot d$. It is a beautiful property of symmetry that holds true regardless of the starting term or the common difference.

Q: Why can't we just use the standard sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$?

You absolutely can! But notice that using the standard formula introduces $a_1$ into your equations, which you then have to eliminate. The pairing method is a 'JEE-style' optimization—it bypasses the need to find $a_1$ entirely, saving you precious minutes and reducing the chance of calculation errors.

Question

Suppose that the number of terms in an A.P. is 2k, $k\in N.$ If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to:

Options:

(A) 6

(B) 5

(C) 8

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a sequence of $2k$ terms: $a_1, a_2, a_3, \dots, a_{2k}$. We are told there are $2k$ terms in total, which implies perfect symmetry.

We have $k$ terms in odd positions ($a_1, a_3, \dots, a_{2k-1}$) and $k$ terms in even positions ($a_2, a_4, \dots, a_{2k}$). The problem provides the sum of the odd terms as $40$ and the sum of the even terms as $55$.

Most students immediately reach for the sum formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

While that is correct, it is the long road. Let us take the path of the mathematician.

### The Pairing Trick

What happens if we subtract the sum of the odd terms from the sum of the even terms? We have $S_{even} - S_{odd} = 55 - 40 = 15$.

Now, look at the terms themselves. If we pair them up:

$$(a_2 - a_1) + (a_4 - a_3) + \dots + (a_{2k} - a_{2k-1}) = 15$$

By the definition of an A.P., the difference between any consecutive term and its predecessor is the common difference, $d$. Since we have $k$ such pairs, we are essentially adding $d$ to itself $k$ times.

Thus, we arrive at the beautiful, compact equation:

$$kd = 15$$

### The Last Term Constraint

The problem also states that the last term exceeds the first term by $27$. Mathematically, this is $a_{2k} - a_1 = 27$.

Using the general term formula $a_n = a_1 + (n-1)d$, we express the last term as $a_{2k} = a_1 + (2k-1)d$. Substituting this into our constraint:

$$(a_1 + (2k-1)d) - a_1 = 27$$

The $a_1$ terms cancel out, leaving us with $(2k-1)d = 27$. Expanding this, we get:

$$2kd - d = 27$$

### The Final Victory

We now have a system of two simple equations: $kd = 15$ and $2kd - d = 27$. Since we know $kd = 15$, we substitute this into the second equation:

$$2(15) - d = 27$$

This simplifies to $30 - d = 27$, which yields **$d = 3$**.

Finally, we return to our first equation, $kd = 15$. Substituting $d = 3$, we get $k(3) = 15$, which means **$k = 5$**.

We have arrived at the solution. Notice how we never had to calculate the first term $a_1$ or the individual sums. That is the power of looking for structure before calculating.

Animated Solution for Mathematics - Sequence and Series: Suppose that the number of terms in an A.P. is 2k, $k \in N .$ If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to:

Select Answer:

Visualized Solution

Defining the Sequence

Sum of Odd Terms

Sum of Even Terms

Subtracting the Sums

Pairing the Terms

Finding $k d$

The Last Term Condition

Expanding the Last Term

Simplifying the Equation

Substituting $k d$

Solving for $d$

Final Calculation for $k$

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Pairing Trick

The Last Term Constraint

The Final Victory

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