Arithmetic Sequence and Series

[a] Find the first 4 terms of the sequence.

To find the first four terms, substitute \(n = 1, 2, 3, 4\) into the formula.

\(u_1 = 3(1) + 4 \quad = 3 + 4 \quad = 7\)

\(u_2 = 3(2) + 4 \quad = 6 + 4 \quad = 10\)

\(u_3 = 3(3) + 4 \quad = 9 + 4 \quad = 13\)

\(u_4 = 3(4) + 4 \quad = 12 + 4 \quad = 16\)

Answer: The first four terms are: \(7,\ 10,\ 13,\ 16\)

[b] Find the \(50^\text{th}\) term of the sequence.

We use the formula: \(u_n = 3n + 4\)

Substitute \(n = 50\):

\(u_{50} = 3(50) + 4 \quad = 150 + 4 \quad = 154\)

Answer: The \(50^\text{th}\) term is \(154\)

[c] Find the sum of the first 20 terms of the sequence.

To find the sum of the first \(n\) terms of an arithmetic sequence, use:  \(S_n = \frac{n}{2}(u_1 + u_n)\)

We already found:  \(u_1 = 7\)

Now calculate \(u_{20}\): \(u_{20} = 3(20) + 4 \quad = 60 + 4 \quad = 64\)

Now substitute into the sum formula:

\(S_{20} = \frac{20}{2}(7 + 64)\)

\(\quad = 10 \cdot 71\)

\(\quad = 710\)

Answer: The sum of the first 20 terms is \(710\)

[a] Find the first term and the common difference.

To find the first term, substitute \(n = 1\):

\(u_1 = 11 – 4(1)\)

\(\quad = 11 – 4\)

\(\quad = 7\)

To find the common difference \(d\), subtract \(u_2 – u_1\):

\(u_2 = 11 – 4(2) \)

\(\quad = 11 – 8\)

\(\quad = 3\)

\(d = u_2 – u_1\)

\(\quad = 3 – 7\)

\(\quad = -4\)

Answer:  First term \(u_1 = 7\).  Common difference \(d = -4\)

[b] Find the \(20^\text{th}\) term.

Use the general term formula: \(u_{20} = 11 – 4(20) \quad = 11 – 80 \quad = -69\)

Answer: The \(20^\text{th}\) term is \(-69\)

[c] Which term of the sequence is \(-109\)?

We solve for \(n\) such that \(u_n = -109\):

\(u_n = 11 – 4n\)

\(-109 = 11 – 4n\)

\(-109 – 11 = -4n\)

\(-120 = -4n\)

\(n = \frac{-120}{-4} = 30\)

Answer: The \(30^\text{th}\) term is \(-109\)

[d] Find the sum of the first 25 terms.

We use the formula: \(S_n = \frac{n}{2}(2u_1 + (n – 1)d)\)

Substitute \(n = 25\), \(u_1 = 7\), \(d = -4\):

\(S_{25} = \frac{25}{2} \left[2(7) + (25 – 1)(-4)\right]\)

\(\quad = \frac{25}{2} \left[14 + 24(-4)\right]\)

\(\quad = \frac{25}{2} \left[14 – 96\right]\)

\(\quad = \frac{25}{2} (-82)\)

\(\quad = 25 \cdot (-41) \quad = -1025\)

Answer: The sum of the first 25 terms is \(-1025\)

[a] Show that it is an arithmetic sequence

We check the common difference between consecutive terms:

\(u_2 – u_1 \quad = \quad 6 – 2 \quad = \quad 4\)

\(u_3 – u_2 \quad = \quad 10 – 6 \quad = \quad 4\)

\(u_4 – u_3 \quad = \quad 14 – 10 \quad = \quad 4\)

Since the difference between consecutive terms is constant (\(d = 4\)), the sequence is arithmetic.

[b] Find the next three terms

Continue the pattern by adding the common difference:

\(u_5 \quad = \quad 14 + 4 \quad = \quad 18\)

\(u_6 \quad = \quad 18 + 4 \quad = \quad 22\)

\(u_7 \quad = \quad 22 + 4 \quad = \quad 26\)

So, the next three terms are: 18, 22, 26.

[c] Find the expression for the \(n^\text{th}\) term

We use the formula for the general term of an arithmetic sequence:

\(u_n \quad = \quad u_1 + (n – 1)d\)

\(u_n \quad = \quad 2 + (n – 1) \cdot 4\)

\(u_n \quad = \quad 2 + 4n – 4\)

\(u_n \quad = \quad 4n – 2\)

So, the general term is \(u_n = 4n – 2\)

[d] Find the \(15^\text{th}\) term of the sequence

Substitute \(n = 15\) into the formula:

\(u_{15}  =  4(15) – 2\)

\(u_{15}  =  60 – 2\)

\(u_{15}  =  58\)

The \(15^\text{th}\) term of the sequence is \(58\).

[a] Find the common difference

\(d \quad = \quad u_2 – u_1\)

\(\quad = \quad 17 – 20\)

\(\quad = \quad -3\)

So, the common difference is \(-3\).

[b] Find the \(10^\text{th}\) term of the sequence

Use the general term formula:

\(u_n \quad = \quad u_1 + (n – 1)d\)

\(u_{10} \quad = \quad 20 + (10 – 1)(-3)\)

\(\quad = \quad 20 + 9(-3)\)

\(\quad = \quad 20 – 27\)

\(\quad = \quad -7\)

The \(10^\text{th}\) term is \(-7\).

[c] Given \(u_n = -37\), find \(n\)

\(u_n \quad = \quad 20 + (n – 1)(-3)\)

\(-37 \quad = \quad 20 – 3(n – 1)\)

\(-37 \quad = \quad 20 – 3n + 3\)

\(-37 \quad = \quad 23 – 3n\)

\(-37 – 23 \quad = \quad -3n\)

\(-60 \quad = \quad -3n\)

\(n \quad = \quad \frac{60}{3}\)

\(n \quad = \quad 20\)

So, \(u_n = -37\) occurs at \(n = 20\).

[d] Find the sum of all positive terms in the sequence

We list the terms until they become non-positive:

\(u_1 = 20\)

\(u_2 = 17\)

\(u_3 = 14\)

\(u_4 = 11\)

\(u_5 = 8\)

\(u_6 = 5\)

\(u_7 = 2\)

\(u_8 = -1\) (stop here)

So the last positive term is \(u_7 = 2\).

Now calculate the sum of the first 7 terms:

\(S_n \quad = \quad \frac{n}{2}(2u_1 + (n – 1)d)\)

\(S_7 \quad = \quad \frac{7}{2}[2(20) + (7 – 1)(-3)]\)

\(\quad = \quad \frac{7}{2}[40 + 6(-3)]\)

\(\quad = \quad \frac{7}{2}[40 – 18]\)

\(\quad = \quad \frac{7}{2}[22]\)

\(\quad = \quad \frac{154}{2}\)

\(\quad = \quad 77\)

The sum of all positive terms is \(77\).

[a] Find the first 4 terms of the sequence

We are given:

• \(u_{16} = 44\)
• \(d = 3\)

Use the formula for the \(n^\text{th}\) term: \(u_n \quad = \quad u_1 + (n – 1)d\)

Substitute to find \(u_1\):

\(u_{16}  =  u_1 + (16 – 1)(3)\)
\(44  =  u_1 + 15 \cdot 3\)
\(44  =  u_1 + 45\)
\(u_1  =  44 – 45\)
\(u_1  = -1\)

Now find the next three terms:

\(u_2  = u_1 + d \quad = \quad -1 + 3 \quad = \quad 2\)
\(u_3  =  u_2 + d \quad = \quad 2 + 3 \quad = \quad 5\)
\(u_4  =  u_3 + d \quad = \quad 5 + 3 \quad = \quad 8\)

So, the first 4 terms are: \(-1,\ 2,\ 5,\ 8\)

[b] Find the \(51^\text{st}\) term

\(u_{51}  =  u_1 + (51 – 1)d\)
\(\quad =  -1 + 50 \cdot 3\)
\(\quad =  -1 + 150\)
\(\quad =  149\)

So, the \(51^\text{st}\) term is \(149\)

[c] Find the sum of the first 20 terms

Use the sum formula: \(S_n =  \frac{n}{2}\left[2u_1 + (n – 1)d\right]\)

\(S_{20} =  \frac{20}{2}\left[2(-1) + (20 – 1)(3)\right]\)

     \(\quad =  10 \cdot [-2 + 57]\)

     \(\quad =  10 \cdot 55\)

      \(\quad = 550\)

The sum of the first 20 terms is \(550\)

We use the general formula of the \(n^\text{th}\) term of an arithmetic sequence: \(u_n = u_1 + (n – 1)d\)

[a] Finding \(u_1\) and \(d\)

We are given:

\(u_{10} = u_1 + 9d = 43\)

\(u_{33} = u_1 + 32d = 204\)

Now subtract the two equations:

\((u_1 + 32d) – (u_1 + 9d) = 204 – 43\)

\(u_1 + 32d – u_1 – 9d = 161\)

\(23d = 161\)

\(d = \frac{161}{23} = 7\)

Substitute \(d = 7\) into \(u_{10} = u_1 + 9d = 43\)

\(u_1 + 9(7) = 43\)

\(u_1 + 63 = 43\)

\(u_1 = 43 – 63\)

\(u_1 = -20\)

So, \(u_1 = -20 \quad \text{and} \quad d = 7\)

[b] Finding \(S_{20}\)

Use the formula: \(S_n = \frac{n}{2}[2u_1 + (n – 1)d]\)

Substitute values:

\(S_{20} = \frac{20}{2}[2(-20) + 19(7)]\)

\(= 10[-40 + 133]\)

\(= 10 \cdot 93\)

\(= 930\)

Final Answers:  \(u_1 = -20\), \(d = 7\), \(S_{20} = 930\)

This is an arithmetic series with:

First term \(u_1 = 11\)

Common difference \(d = 18 – 11 = 7\)

Last term \(u_n = 74\)

We use the \(n^\text{th}\) term formula to find \(n\): \(u_n = u_1 + (n – 1)d\)

\(74 = 11 + (n – 1)(7)\)

\(74 = 11 + 7n – 7\)

\(74 = 7n + 4\)

\(70 = 7n\)

\(n = 10\)

Now use the sum formula:

\(S_n = \frac{n}{2}(u_1 + u_n)\)

\(S_{10} = \frac{10}{2}(11 + 74)\)

\(= 5 \cdot 85\)

\(= 425\)

Final Answer: \(\textbf{425}\)

We are given:

• • • \(u_1 = 3\)
    • \(u_{25} = 51\)

First, use the general term formula to find the common difference \(d\): \(u_n = u_1 + (n – 1)d\)

Substitute \(n = 25\):

\(51 = 3 + (25 – 1)d\)

\(51 = 3 + 24d\)

\(48 = 24d\)

\(d = 2\)

Now, use the sum formula to find \(S_{30}\):

\(S_n = \frac{n}{2}[2u_1 + (n – 1)d]\)

Substitute the values:

\(S_{30} = \frac{30}{2}[2(3) + (30 – 1)(2)]\)

\(S_{30} = 15[6 + 58]\)

\(S_{30} = 15 \cdot 64\)

\(S_{30} = 960\)

\(\textbf{Final Answer:} \quad 960\)

We are given:

• \(u_1 = 16\)
• \(u_n = 81\)
• \(S_n = 679\)

Use the formula for the sum of the first \(n\) terms:   \(S_n = \frac{n}{2}(u_1 + u_n)\)

Substitute known values:

\(679 = \frac{n}{2}(16 + 81)\)

\(679 = \frac{n}{2}(97)\)

\(679 = \frac{97n}{2}\)

Multiply both sides by 2:

\(1358 = 97n\)

Solve for \(n\):

\(n = \frac{1358}{97}\)

\(n = 14\)

\(\textbf{Final Answer:} \quad 14\)

We use the general term formula: \(u_n = u_1 + (n – 1)d\)

[a] Use two equations to find \(u_1\) and \(d\)

\(u_{10} = u_1 + 9d = 109 \quad \text{(1)}\)

\(u_{38} = u_1 + 37d = 389 \quad \text{(2)}\)

Subtract (1) from (2):

\((u_1 + 37d) – (u_1 + 9d) = 389 – 109\)

\(u_1 + 37d – u_1 – 9d = 280\)

\(28d = 280\)

\(d = 10\)

Substitute \(d = 10\) into equation (1):

\(u_1 + 9(10) = 109\)

\(u_1 + 90 = 109\)

\(u_1 = 19\)

[b] Find the \(20^\text{th}\) term

\(u_{20} = u_1 + 19d\)

\(u_{20} = 19 + 19 \cdot 10\)

\(u_{20} = 19 + 190\)

\(u_{20} = 209\)

\(\textbf{Final Answers:}\) \(\quad u_1 = 19\) ; \(\quad d = 10\) ; \(\quad u_{20} = 209\)

We use the general term formula: \(u_n = u_1 + (n – 1)d\)

Step 1: Use the two terms to form equations

From \(u_{14} = -52\):  \(u_1 + 13d = -52 \quad \text{(1)}\)

From \(u_{40} = -156\):  \(u_1 + 39d = -156 \quad \text{(2)}\)

Step 2: Subtract the equations

\((u_1 + 39d) – (u_1 + 13d) = -156 – (-52)\)

\(u_1 + 39d – u_1 – 13d = -104\)

\(26d = -104\)

\(d = -4\)

Step 3: Substitute back to find \(u_1\)

Substitute \(d = -4\) into (1):

\(u_1 + 13(-4) = -52\)

\(u_1 – 52 = -52\)

\(u_1 = 0\)

Step 4: Find the 50th term \(u_{50}\)

\(u_{50} = u_1 + 49d\)

\(u_{50} = 0 + 49(-4)\)

\(u_{50} = -196\)

Step 5: Use sum formula \(S_n = \frac{n}{2}(u_1 + u_n)\)

\(S_{50} = \frac{50}{2}(0 + (-196))\)

\(S_{50} = 25 \cdot (-196)\)

\(S_{50} = -4900\)

\(\textbf{Final Answer:} \quad \boxed{-4900}\)

We use the formula for the sum of the first \(n\) terms of an arithmetic sequence:
\(S_n = \frac{n}{2}(u_1 + u_n)\)

Substitute the known values:

\(-11610 = \frac{n}{2}(-38 + (-478))\)

\(-11610 = \frac{n}{2}(-516)\)

\(-11610 = -258n\)

Solve for \(n\):

\(n = \frac{-11610}{-258}\)

\(n = 45\)