Consider an A.P.: a1, a2, ..., an; a1 0. If a2 - a1 = -3/4, an = 1/4 a1, and Σ (i=1 to n) ai = 525

Q: Wait, why can't we just use the sum formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$ directly?

You absolutely can! However, using $S_n = \frac{n}{2}(a_1 + a_n)$ is often more efficient when the last term $a_n$ is given as a simple fraction of $a_1$. It reduces the number of variables you need to juggle, making the algebra cleaner and less prone to errors.

Q: How do I know $n$ must be an integer?

In any arithmetic progression, $n$ represents the count of terms. You cannot have a fractional or negative number of terms in a sequence. If your quadratic equation yields a non-integer or negative result for $n$, it is a strong signal to re-check your algebraic steps.

Q: Why did we reject the negative root of the quadratic equation?

Great observation! The quadratic equation $n^2 - n - 420 = 0$ gives two roots: $n = 21$ and $n = -20$. Since $n$ represents the number of terms in a sequence, it must be a positive integer. Therefore, we discard the negative root as it is physically impossible in this context.

Question

Consider an A.P.: $a_1, a_2, \dots, a_n; a_1 > 0$. If $a_2 - a_1 = -3/4, a_n = \frac{1}{4} a_1$, and $\sum_{i=1}^n a_i = 525/2$, then $\sum_{i=1}^{17} a_i$ is equal to

Options:

(A) 136

(B) 476

(C) 238

(D) 952

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given an arithmetic progression (A.P.) starting with a positive term $a_1$ and a common difference $d = -\frac{3}{4}$.

The problem provides a specific relationship between the last term and the first term:
$$a_n = \frac{1}{4} a_1$$

Using the general formula for the $n$-th term, $a_n = a_1 + (n-1)d$, we substitute our known values:
$$\frac{a_1}{4} = a_1 + (n-1)\left(-\frac{3}{4}ight)$$

### The Master Equation

By rearranging the terms to isolate the variables, we obtain:
$$a_1 - \frac{a_1}{4} = (n-1)\left(\frac{3}{4}ight)$$

This simplifies to:
$$\frac{3a_1}{4} = \frac{3(n-1)}{4}$$

Canceling the common factor of $\frac{3}{4}$ from both sides, we arrive at the elegant relation:
$$a_1 = n-1$$

### The Summation Bridge

We are given that the sum of $n$ terms is $S_n = \frac{525}{2}$. Using the sum formula $S_n = \frac{n}{2}(a_1 + a_n)$, we substitute $a_n = \frac{a_1}{4}$:
$$S_n = \frac{n}{2}\left(a_1 + \frac{a_1}{4}ight) = \frac{n}{2}\left(\frac{5a_1}{4}ight)$$

Substituting our earlier finding $a_1 = n-1$ into this equation yields:
$$\frac{525}{2} = \frac{n}{2} \cdot \frac{5(n-1)}{4}$$

This simplifies to:
$$\frac{525}{2} = \frac{5n(n-1)}{8}$$

### Solving the Quadratic

Clearing the fractions, we calculate:
$$n(n-1) = \frac{525 	imes 8}{2 	imes 5} = 105 	imes 4 = 420$$

This results in the quadratic equation:
$$n^2 - n - 420 = 0$$

Factoring the quadratic, we look for two numbers that multiply to $-420$ and add to $-1$, which are $-21$ and $20$:
$$(n-21)(n+20) = 0$$

Since $n$ must be a positive integer, we find $n = 21$. Consequently, the first term is $a_1 = 21 - 1 = 20$.

### Final Calculation

We now have the parameters $a_1 = 20$ and $d = -\frac{3}{4}$. We need to find the sum of the first 17 terms, $S_{17}$.

Using the sum formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$, we substitute our values:
$$S_{17} = \frac{17}{2}\left[2(20) + (17-1)\left(-\frac{3}{4}ight)ight]$$

Simplifying the expression inside the brackets:
$$S_{17} = \frac{17}{2}[40 + 16(-\frac{3}{4})] = \frac{17}{2}[40 - 12] = \frac{17}{2} 	imes 28$$

Performing the final multiplication:
$$S_{17} = 17 	imes 14 = 238$$

The final result is **238**.

Animated Solution for Mathematics - Sequence and Series: Consider an A.P.: $a_{1}, a_{2}, \dots, a_{n}; a_{1} > 0$ . If $a_{2} - a_{1} = - 3/4, a_{n} = \frac{1}{4} a_{1}$ , and $\sum_{i = 1}^{n} a_{i} = 525/2$ , then $\sum_{i = 1}^{17} a_{i}$ is equal to

Select Answer:

Visualized Solution

Understanding the A.P.

Relating $a_{n}$ and $a_{1}$

Substituting Known Values

Simplifying the Equation

Expressing $a_{1}$ in terms of $n$

The Sum of $n$ Terms

Substituting into the Sum Formula

Forming a Quadratic in $n$

Simplifying to a Standard Quadratic

Solving for $n$

Finding the First Term $a_{1}$

Setting up the Target Sum $S_{17}$

Substituting Values into $S_{17}$

Calculating the Final Sum

Final Answer

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

Analyzing the Setup

The Master Equation

The Summation Bridge

Solving the Quadratic

Final Calculation

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Select Answer:

Visualized Solution

Understanding the A.P.

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​

Substituting Known Values

Simplifying the Equation

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​ in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n Terms

Substituting into the Sum Formula

Forming a Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Simplifying to a Standard Quadratic

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Finding the First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​

Setting up the Target Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S17​

Substituting Values into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S17​