Let the sum of the first n terms of a non-constant A.P., a₁, a₂, a₃, ... be 50n + n(n-7)/2 A, where

Q: Why is there no constant term in the sum of an AP?

The sum $S_n$ represents the sum of $n$ terms. If $n=0$, the sum must be 0. If there were a constant term, $S_0$ would be non-zero, which is physically impossible for a sum of zero terms. In any valid AP sum formula, the constant term must be zero.

Q: Can I just find $a_1$ and $a_2$ and then find $d$?

Absolutely! You can find $a_1 = S_1$ and $a_2 = S_2 - S_1$. Then $d = a_2 - a_1$. This is a fantastic way to verify your answer and build confidence in your result.

Q: Why did we use $a_{50} = a_1 + 49d$ instead of $S_{50} - S_{49}$?

Both are correct! $S_{50} - S_{49}$ is the definition of the 50th term. However, using the general term formula $a_n = a_1 + (n-1)d$ is often faster and less prone to calculation errors once you have found $a_1$ and $d$.

Question

Let the sum of the first $n$ terms of a non-constant A.P., $a_1, a_2, a_3, \dots$ be $50n + \frac{n(n-7)}{2}A$, where $A$ is a constant. If $d$ is the common difference of this A.P., then the ordered pair $(d, a_{50})$ is equal to

Options:

(A) $(A, 50+46A)$

(B) $(A, 50+45A)$

(C) $(50, 50+46A)$

(D) $(50, 50+45A)$

Sigma Percentile · Accepted Answer

### Analyzing the Anatomy of a Sum

We are given the sum of the first $n$ terms of a non-constant Arithmetic Progression (AP):

$$S_n = 50n + \frac{n(n-7)}{2}A$$

At first glance, this expression appears complex. However, we must recall the universal standard form for the sum of an AP:

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

Expanding this standard form yields:

$$S_n = \frac{d}{2}n^2 + \left(a - \frac{d}{2}\right)n$$

This is the **Algebraic Mirror**. Any valid sum of an AP must be a quadratic in $n$ with no constant term. If a constant term exists, the sequence is not a pure AP.

### The Algebraic Transformation

To align our given expression with the standard form, we expand the provided equation:

$$S_n = 50n + \frac{A}{2}(n^2 - 7n)$$

Next, we group the terms by the powers of $n$ to isolate the coefficients:

$$S_n = \frac{A}{2}n^2 + \left(50 - \frac{7A}{2}\right)n$$

The chaos has now vanished, revealing a clear quadratic structure. We can now identify the coefficients of $n^2$ and $n$ directly.

### The Coefficient Comparison

By comparing our transformed expression with the standard form $S_n = \frac{d}{2}n^2 + (a - \frac{d}{2})n$, we extract the parameters of the sequence.

Equating the coefficients of $n^2$:

$$\frac{d}{2} = \frac{A}{2} \implies d = A$$

Thus, the common difference is simply $A$. To find the first term $a_1$, we utilize the property $S_1 = a_1$:

$$a_1 = 50(1) + \frac{1(1-7)}{2}A = 50 - 3A$$

### The Final Calculation

We have determined that the common difference is $d = A$ and the first term is $a_1 = 50 - 3A$. The problem requires the 50th term, $a_{50}$, which is calculated using the formula $a_n = a_1 + (n-1)d$:

$$a_{50} = a_1 + 49d$$

Substituting our derived values into this equation:

$$a_{50} = (50 - 3A) + 49(A)$$

$$a_{50} = 50 + 46A$$

The resulting ordered pair $(d, a_{50})$ is **$(A, 50 + 46A)$**. By observing the underlying structure of the sum, we have bypassed complex arithmetic to reach the solution efficiently.

Select Answer:

Visualized Solution

Analyze the Given Sum $S_{n}$

Recall the General Form of $S_{n}$

Expand the Given Expression

Group the Terms of $n^{2}$ and $n$

Determine Common Difference $d$

Find the First Term $a_{1}$

Calculate the 50th Term $a_{50}$

Final Result and Ordered Pair

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

Analyzing the Anatomy of a Sum

The Algebraic Transformation

The Coefficient Comparison

The Final Calculation

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Visualized Solution

Analyze the Given Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Recall the General Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​