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

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

136

476

238

952

The correct answer is

476

AP Sum Calculation: Finding $\sum_{i=1}^{17} a_i$

Given Information

The sequence is an Arithmetic Progression (A.P.): $a_1, a_2, \dots, a_n$.
The first term is positive: $a_1 > 0$.
The common difference is $d = a_2 - a_1 = \frac{-3}{4}$.
The $n$-th term is related to the first term: $a_n = \frac{1}{4}a_1$.
The sum of the first $n$ terms is $S_n = \sum_{i=1}^n a_i = \frac{525}{2}$.

Solution Steps

1. Find relation between $a_1$ and $n$:

Use the formula for the $n$-th term of an A.P.: $a_n = a_1 + (n-1)d$. Substitute the given values $a_n = \frac{1}{4}a_1$ and $d = \frac{-3}{4}$:

$\frac{1}{4}a_1 = a_1 + (n-1)\left(\frac{-3}{4}\right)$

Rearranging the terms to solve for $n-1$:

$\frac{1}{4}a_1 - a_1 = (n-1)\left(\frac{-3}{4}\right)$

$\frac{-3}{4}a_1 = (n-1)\left(\frac{-3}{4}\right)$

Since $a_1 > 0$, we can divide by $\frac{-3}{4}$, which gives:

$a_1 = n-1$

This means $n = a_1 + 1$.

2. Determine $a_1$ and $n$:

Use the formula for the sum of the first $n$ terms: $S_n = \frac{n}{2}(a_1 + a_n)$. Substitute $S_n = \frac{525}{2}$, $a_n = \frac{1}{4}a_1$, and the relation $n = a_1 + 1$:

$\frac{525}{2} = \frac{a_1 + 1}{2}\left(a_1 + \frac{1}{4}a_1\right)$

Simplify the expression:

$\frac{525}{2} = \frac{a_1 + 1}{2}\left(\frac{5}{4}a_1\right)$

Multiply both sides by 2:

$525 = (a_1 + 1)\left(\frac{5}{4}a_1\right)$

Isolate the terms involving $a_1$:

$525 \times \frac{4}{5} = a_1(a_1 + 1)$

$420 = a_1^2 + a_1$

Form the quadratic equation:

$a_1^2 + a_1 - 420 = 0$

Factor the quadratic equation: $(a_1 + 21)(a_1 - 20) = 0$.

Since $a_1 > 0$, the valid solution is $a_1 = 20$.

Calculate $n$ using $n = a_1 + 1$:

$n = 20 + 1 = 21$

The parameters are $a_1 = 20$, $d = \frac{-3}{4}$, and $n = 21$.

3. Calculate the sum of the first 17 terms ($\sum_{i=1}^{17} a_i$):

Use the A.P. sum formula $S_{17} = \frac{17}{2}(2a_1 + (17-1)d)$.

Substitute the values $a_1 = 20$ and $d = \frac{-3}{4}$:

$S_{17} = \frac{17}{2}\left(2(20) + 16\left(\frac{-3}{4}\right)\right)$

Calculate the terms inside the parenthesis:

$S_{17} = \frac{17}{2}\left(40 - 12\right)$

$S_{17} = \frac{17}{2}(28)$

Perform the final multiplication:

$S_{17} = 17 \times 14 = 238$

Result

The sum of the first 17 terms is 238.

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