50+ Math worksheets for 12th Grade on Quizizz | Free & Printable - Free Printable

Let's solve each of these Pre-Calculus Sequences and Series problems step by step.

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1. Find the common difference: $ a_2 = 6 $ and $ a_6 = -14 $

This is an arithmetic sequence, so we use the formula:

$$
a_n = a_1 + (n-1)d
$$

We know:
- $ a_2 = a_1 + d = 6 $
- $ a_6 = a_1 + 5d = -14 $

Subtract the first equation from the second:

$$
(a_1 + 5d) - (a_1 + d) = -14 - 6 \\
4d = -20 \Rightarrow d = -5
$$

✔ Answer: A. -5

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2. Find the common ratio (multiplier): $ a_1 = 4 $, $ a_4 = 256 $

This is a geometric sequence:

$$
a_n = a_1 \cdot r^{n-1}
$$

So:
$$
a_4 = 4 \cdot r^3 = 256 \\
r^3 = \frac{256}{4} = 64 \\
r = \sqrt[3]{64} = 4
$$

✔ Answer: C. 4

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3. Write the nth term formula: $ 5 + 11 + 17 + 23 + \ldots + 35 $

Check if it's arithmetic:
- $ 11 - 5 = 6 $
- $ 17 - 11 = 6 $ → Common difference $ d = 6 $
- First term $ a = 5 $

Formula for arithmetic sequence:
$$
t(n) = a + (n-1)d = 5 + (n-1)(6)
$$

So:
$$
t(n) = 5 + 6(n-1)
$$

✔ Answer: A. $ t(n) = 5 + 6(n-1) $

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4. Write the nth term formula: $ 3.5 + 7 + 14 + 28 + 56 $

Check the pattern:
- $ 7 / 3.5 = 2 $
- $ 14 / 7 = 2 $
- $ 28 / 14 = 2 $ → This is a geometric sequence with common ratio $ r = 2 $

First term $ a = 3.5 $

Geometric formula:
$$
a_n = a \cdot r^{n-1} = 3.5 \cdot (2)^{n-1}
$$

✔ Answer: D. $ a_n = 3.5(2)^{n-1} $

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5. Find the sum: $ 3 + 8 + 13 + 18 + \ldots + 98 $

Arithmetic series:
- $ a = 3 $
- $ d = 5 $
- Last term $ l = 98 $

Find number of terms $ n $:

$$
a_n = a + (n-1)d = 98 \\
3 + (n-1)(5) = 98 \\
(n-1)(5) = 95 \\
n-1 = 19 \Rightarrow n = 20
$$

Sum formula:
$$
S_n = \frac{n}{2}(a + l) = \frac{20}{2}(3 + 98) = 10 \cdot 101 = 1010
$$

✔ Answer: C. 1010

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6. Find the sum: $ 11 + 12 + 13 + \ldots + 40 $

This is an arithmetic series:
- $ a = 11 $
- $ l = 40 $
- $ d = 1 $

Number of terms:
$$
n = 40 - 11 + 1 = 30
$$

Sum:
$$
S_n = \frac{n}{2}(a + l) = \frac{30}{2}(11 + 40) = 15 \cdot 51 = 765
$$

✔ Answer: B. 765

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7. Find the sum: $ 90 + 84 + 78 + \ldots + 24 $

Arithmetic series:
- $ a = 90 $
- $ d = -6 $
- $ l = 24 $

Find number of terms:
$$
a_n = a + (n-1)d = 24 \\
90 + (n-1)(-6) = 24 \\
(n-1)(-6) = -66 \\
n-1 = 11 \Rightarrow n = 12
$$

Sum:
$$
S_n = \frac{n}{2}(a + l) = \frac{12}{2}(90 + 24) = 6 \cdot 114 = 684
$$

But wait — this isn't one of the options? Let's check the options again:

> A. ?
> B. ?
> C. ?
> D. ?

Wait — the image shows only A, B, C, D, but no numbers listed for question 7. But let’s recheck.

Wait — in your image, question 7 says:
"Find the sum: 90 + 84 + 78 + ... + 24"

We computed:
- $ n = 12 $
- $ S = 6 \times 114 = 684 $

But none of the options are shown in the image — perhaps they're missing?

Wait, maybe you didn’t include the choices for Q7.

But based on calculation:

✔ Sum = 684

If the choices were:
A. 684
B. 690
C. 700
D. 720

Then answer would be A. 684

But since the image doesn't show them, I’ll assume the problem was just to compute it.

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✔ Final Answers Summary:

| Question | Answer |
|--------|--------|
| 1 | A. -5 |
| 2 | C. 4 |
| 3 | A. $ t(n) = 5 + 6(n-1) $ |
| 4 | D. $ a_n = 3.5(2)^{n-1} $ |
| 5 | C. 1010 |
| 6 | B. 765 |
| 7 | 684 (not listed — likely missing options) |

Let me know if you want me to help with the remaining questions or clarify anything!