2. (a) Determine $$ n $$, if... (i) $$ \sum_{r=1}^{n}(2 r-3)=840 $$ (ii) $$ \sum_{k=1}^{n}(3 k-7)=52 $$ (b) Determine $$ k $$, if... (i) $$ \sum_{t=2}^{k}(6 t-1)=795 $$ (ii) $$ \sum_{i=0}^{k}(1-3 i)=-125 $$

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UpStudy AI · Accepted Answer

Let's solve each part of the problem step by step.

### 2. (a) Determine $$ n $$, if...

#### (i) $$ \sum_{r=1}^{n}(2 r - 3) = 840 $$

**Solution:**

Compute the sum:
$$
\sum_{r=1}^{n} (2r - 3) = 2\sum_{r=1}^{n} r - 3\sum_{r=1}^{n} 1 = 2 \cdot \frac{n(n+1)}{2} - 3n = n(n+1) - 3n = n^2 - 2n
$$
Set the sum equal to 840:
$$
n^2 - 2n = 840 \
n^2 - 2n - 840 = 0
$$
Solve the quadratic equation:
$$
n = \frac{2 \pm \sqrt{4 + 3360}}{2} = \frac{2 \pm \sqrt{3364}}{2} = \frac{2 \pm 58}{2}
$$
Discard the negative solution:
$$
n = \frac{2 + 58}{2} = 30
$$

**Answer:** $$ n = 30 $$

#### (ii) $$ \sum_{k=1}^{n}(3 k - 7) = 52 $$

**Solution:**

Compute the sum:
$$
\sum_{k=1}^{n} (3k - 7) = 3\sum_{k=1}^{n} k - 7n = 3 \cdot \frac{n(n+1)}{2} - 7n = \frac{3n^2 + 3n}{2} - 7n = \frac{3n^2 - 11n}{2}
$$
Set the sum equal to 52:
$$
\frac{3n^2 - 11n}{2} = 52 \
3n^2 - 11n - 104 = 0
$$
Solve the quadratic equation:
$$
n = \frac{11 \pm \sqrt{121 + 1248}}{6} = \frac{11 \pm \sqrt{1369}}{6} = \frac{11 \pm 37}{6}
$$
Discard the negative solution:
$$
n = \frac{11 + 37}{6} = 8
$$

**Answer:** $$ n = 8 $$

### 2. (b) Determine $$ k $$, if...

#### (i) $$ \sum_{t=2}^{k}(6 t - 1) = 795 $$

**Solution:**

Compute the sum from $$ t=1 $$ to $$ k $$ and subtract the $$ t=1 $$ term:
$$
\sum_{t=1}^{k} (6t - 1) = 6 \cdot \frac{k(k+1)}{2} - k = 3k(k+1) - k = 3k^2 + 2k
$$
Subtract the $$ t=1 $$ term ($$ 6(1) - 1 = 5 $$):
$$
\sum_{t=2}^{k} (6t - 1) = 3k^2 + 2k - 5 = 795 \
3k^2 + 2k - 800 = 0
$$
Solve the quadratic equation:
$$
k = \frac{-2 \pm \sqrt{4 + 9600}}{6} = \frac{-2 \pm \sqrt{9604}}{6} = \frac{-2 \pm 98}{6}
$$
Discard the negative solution:
$$
k = \frac{-2 + 98}{6} = 16
$$

**Answer:** $$ k = 16 $$

#### (ii) $$ \sum_{i=0}^{k}(1 - 3i) = -125 $$

**Solution:**

Compute the sum:
$$
\sum_{i=0}^{k} (1 - 3i) = \sum_{i=0}^{k} 1 - 3\sum_{i=0}^{k} i = (k+1) - 3 \cdot \frac{k(k+1)}{2} = (k+1) - \frac{3k(k+1)}{2}
$$
Set the sum equal to -125:
$$
k + 1 - \frac{3k(k+1)}{2} = -125
$$
Multiply both sides by 2 to eliminate the fraction:
$$
2(k + 1) - 3k(k + 1) = -250 \
2k + 2 - 3k^2 - 3k = -250 \
-3k^2 - k + 2 = -250 \
-3k^2 - k + 252 = 0 \
3k^2 + k - 252 = 0
$$
Solve the quadratic equation:
$$
k = \frac{-1 \pm \sqrt{1 + 3024}}{6} = \frac{-1 \pm \sqrt{3025}}{6} = \frac{-1 \pm 55}{6}
$$
Discard the negative solution:
$$
k = \frac{-1 + 55}{6} = 9
$$

**Answer:** $$ k = 9 $$

### Summary of Answers

- **2(a)(i)**: $$ n = 30 $$
- **2(a)(ii)**: $$ n = 8 $$
- **2(b)(i)**: $$ k = 16 $$
- **2(b)(ii)**: $$ k = 9 $$
$$ n = 30 $$, $$ n = 8 $$, $$ k = 16 $$, $$ k = 9 $$.

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