Identify the sequence:

• First term: $a_1 = 5$
• Common difference: $d = 9 - 5 = 4$

Explicit formula: $a_n = a_1 + (n-1)d$ $a_n = 5 + (n-1)(4)$ $a_n = 5 + 4n - 4$ $a_n = 4n + 1$

Recursive formula: $a_n = a_{n-1} + 4, \quad a_1 = 5$

Find the 20th term: $a_{20} = 4(20) + 1 = 80 + 1 = 81$

Verify: We can check by adding $d$ nineteen times to $a_1$: $5 + 19(4) = 5 + 76 = 81$ ✓

Answers:

• Explicit: $a_n = 4n + 1$
• Recursive: $a_n = a_{n-1} + 4, a_1 = 5$
• 20th term: $81$

Solution:

Part (a): For arithmetic sequence: $a_n = a_1 + (n-1)d$

$a_1 = 5$ $a_2 = 5 + 3 = 8$ $a_3 = 8 + 3 = 11$ $a_4 = 11 + 3 = 14$ $a_5 = 14 + 3 = 17$

First five terms: 5, 8, 11, 14, 17

Part (b): $a_{20} = a_1 + (20-1)d = 5 + 19(3) = 5 + 57 = 62$

Part (c): Sum formula: $S_n = \frac{n(a_1 + a_n)}{2}$

$S_{20} = \frac{20(5 + 62)}{2} = \frac{20(67)}{2} = 10(67) = 670$

Given information:

• $a_1 = 3$
• $r = 2$

Find the 8th term using explicit formula: $a_n = a_1 \cdot r^{n-1}$ $a_8 = 3 \cdot 2^{8-1}$ $a_8 = 3 \cdot 2^7$ $a_8 = 3 \cdot 128 = 384$

Find the sum of first 8 terms: $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$ $S_8 = 3 \cdot \frac{1 - 2^8}{1 - 2}$ $S_8 = 3 \cdot \frac{1 - 256}{-1}$ $S_8 = 3 \cdot \frac{-255}{-1}$ $S_8 = 3 \cdot 255 = 765$

Verify the sequence: $3, 6, 12, 24, 48, 96, 192, 384$ Sum: $3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765$ ✓

Answers:

• 8th term: $384$
• Sum of first 8 terms: $765$

Solution:

Part (a): For geometric sequence: $a_n = a_1 \cdot r^{n-1}$

$a_6 = 2 \cdot 3^{6-1} = 2 \cdot 3^5 = 2 \cdot 243 = 486$

Part (b): Sum formula: $S_n = a_1 \cdot \frac{r^n - 1}{r - 1}$ (for $r \neq 1$)

$S_8 = 2 \cdot \frac{3^8 - 1}{3 - 1}$

$= 2 \cdot \frac{6561 - 1}{2}$

$= 2 \cdot \frac{6560}{2}$

$= 2 \cdot 3280$

$= 6560$

Set up equations using $a_n = a_1 + (n-1)d$:

For the 3rd term: $a_3 = a_1 + 2d = 14$

For the 7th term: $a_7 = a_1 + 6d = 30$

Solve the system by elimination:

Subtract first equation from second: $(a_1 + 6d) - (a_1 + 2d) = 30 - 14$ $4d = 16$ $d = 4$

Find $a_1$: Substitute $d = 4$ into first equation: $a_1 + 2(4) = 14$ $a_1 + 8 = 14$ $a_1 = 6$

Write the explicit formula: $a_n = a_1 + (n-1)d$ $a_n = 6 + (n-1)(4)$ $a_n = 6 + 4n - 4$ $a_n = 4n + 2$

Verify:

• $a_3 = 4(3) + 2 = 14$ ✓
• $a_7 = 4(7) + 2 = 30$ ✓

Answers:

• First term: $a_1 = 6$
• Common difference: $d = 4$
• Explicit formula: $a_n = 4n + 2$