7. Find an expression for the $$ n $$th term of each of the quadratic sequences: $$ \begin{array}{lll} ext { (i) } 7,10,15,22,31, \ldots & ext { (ii) } 2,7,14,23, \ldots & ext { (iii) } 2,8,18,32,50, \ldots\end{array} $$ 8. Show that the $$ n $$th term of the quadratic sequence $$ 8,15,26,41,60, \ldots $$ is given by $$ T_{n}=2 n^{2}+n+5 $$

Question

7. Find an expression for the $$ n $$th term of each of the quadratic sequences: $$ \begin{array}{lll}	ext { (i) } 7,10,15,22,31, \ldots & 	ext { (ii) } 2,7,14,23, \ldots & 	ext { (iii) } 2,8,18,32,50, \ldots\end{array} $$ 8. Show that the $$ n $$th term of the quadratic sequence $$ 8,15,26,41,60, \ldots $$ is given by $$ T_{n}=2 n^{2}+n+5 $$

UpStudy AI · Accepted Answer

Sure, let's tackle both problems step by step.

---

## **Problem 7**

**Find an expression for the $$ n $$th term of each of the quadratic sequences:**

### **(i) Sequence:** $$ 7, 10, 15, 22, 31, \ldots $$

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  10 - 7 = 3,\quad 15 - 10 = 5,\quad 22 - 15 = 7,\quad 31 - 22 = 9,\ldots
  $$
  Sequence: $$ 3, 5, 7, 9, \ldots $$

- **Second Differences:**
  $$
  5 - 3 = 2,\quad 7 - 5 = 2,\quad 9 - 7 = 2,\ldots
  $$
  The second differences are constant ($$ 2 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

For a quadratic sequence, the general term is:
$$
T_n = an^2 + bn + c
$$
Given that the second difference is $$ 2 $$, we know:
$$
2a = 2 \quad \Rightarrow \quad a = 1
$$
So, the general form simplifies to:
$$
T_n = n^2 + bn + c
$$
Using the first term ($$ n = 1 $$):
$$
7 = 1^2 + b(1) + c \quad \Rightarrow \quad b + c = 6 \quad \text{(Equation 1)}
$$
Using the second term ($$ n = 2 $$):
$$
10 = 2^2 + b(2) + c \quad \Rightarrow \quad 4 + 2b + c = 10 \quad \Rightarrow \quad 2b + c = 6 \quad \text{(Equation 2)}
$$
Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 6 - 6 \quad \Rightarrow \quad b = 0
$$
Substitute $$ b = 0 $$ into Equation 1:
$$
0 + c = 6 \quad \Rightarrow \quad c = 6
$$
**Final Formula:**
$$
T_n = n^2 + 6
$$

---

### **(ii) Sequence:** $$ 2, 7, 14, 23, \ldots $$

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  7 - 2 = 5,\quad 14 - 7 = 7,\quad 23 - 14 = 9,\ldots
  $$
  Sequence: $$ 5, 7, 9, \ldots $$

- **Second Differences:**
  $$
  7 - 5 = 2,\quad 9 - 7 = 2,\ldots
  $$
  The second differences are constant ($$ 2 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

$$
T_n = an^2 + bn + c
$$
With $$ 2a = 2 $$, we have $$ a = 1 $$:
$$
T_n = n^2 + bn + c
$$
Using the first term ($$ n = 1 $$):
$$
2 = 1^2 + b(1) + c \quad \Rightarrow \quad b + c = 1 \quad \text{(Equation 1)}
$$
Using the second term ($$ n = 2 $$):
$$
7 = 2^2 + b(2) + c \quad \Rightarrow \quad 4 + 2b + c = 7 \quad \Rightarrow \quad 2b + c = 3 \quad \text{(Equation 2)}
$$
Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 3 - 1 \quad \Rightarrow \quad b = 2
$$
Substitute $$ b = 2 $$ into Equation 1:
$$
2 + c = 1 \quad \Rightarrow \quad c = -1
$$
**Final Formula:**
$$
T_n = n^2 + 2n - 1
$$

---

### **(iii) Sequence:** $$ 2, 8, 18, 32, 50, \ldots $$

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  8 - 2 = 6,\quad 18 - 8 = 10,\quad 32 - 18 = 14,\quad 50 - 32 = 18,\ldots
  $$
  Sequence: $$ 6, 10, 14, 18, \ldots $$

- **Second Differences:**
  $$
  10 - 6 = 4,\quad 14 - 10 = 4,\quad 18 - 14 = 4,\ldots
  $$
  The second differences are constant ($$ 4 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

$$
T_n = an^2 + bn + c
$$
With $$ 2a = 4 $$, we have $$ a = 2 $$:
$$
T_n = 2n^2 + bn + c
$$
Using the first term ($$ n = 1 $$):
$$
2 = 2(1)^2 + b(1) + c \quad \Rightarrow \quad 2 + b + c = 2 \quad \Rightarrow \quad b + c = 0 \quad \text{(Equation 1)}
$$
Using the second term ($$ n = 2 $$):
$$
8 = 2(2)^2 + b(2) + c \quad \Rightarrow \quad 8 + 2b + c = 8 \quad \Rightarrow \quad 2b + c = 0 \quad \text{(Equation 2)}
$$
Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 0 - 0 \quad \Rightarrow \quad b = 0
$$
Substitute $$ b = 0 $$ into Equation 1:
$$
0 + c = 0 \quad \Rightarrow \quad c = 0
$$
**Final Formula:**
$$
T_n = 2n^2
$$

---

## **Problem 8**

**Show that the $$ n $$th term of the quadratic sequence $$ 8, 15, 26, 41, 60, \ldots $$ is given by $$ T_{n}=2 n^{2}+n+5 $$.**

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  15 - 8 = 7,\quad 26 - 15 = 11,\quad 41 - 26 = 15,\quad 60 - 41 = 19,\ldots
  $$
  Sequence: $$ 7, 11, 15, 19, \ldots $$

- **Second Differences:**
  $$
  11 - 7 = 4,\quad 15 - 11 = 4,\quad 19 - 15 = 4,\ldots
  $$
  The second differences are constant ($$ 4 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

Assume the general term is:
$$
T_n = an^2 + bn + c
$$
Given that the second difference is $$ 4 $$, we have:
$$
2a = 4 \quad \Rightarrow \quad a = 2
$$
So, the general form becomes:
$$
T_n = 2n^2 + bn + c
$$

**Step 3: Use Given Terms to Find $$ b $$ and $$ c $$**

Using the first three terms:

1. For $$ n = 1 $$:
   $$
   8 = 2(1)^2 + b(1) + c \quad \Rightarrow \quad 2 + b + c = 8 \quad \Rightarrow \quad b + c = 6 \quad \text{(Equation 1)}
   $$

2. For $$ n = 2 $$:
   $$
   15 = 2(2)^2 + b(2) + c \quad \Rightarrow \quad 8 + 2b + c = 15 \quad \Rightarrow \quad 2b + c = 7 \quad \text{(Equation 2)}
   $$

3. For $$ n = 3 $$:
   $$
   26 = 2(3)^2 + b(3) + c \quad \Rightarrow \quad 18 + 3b + c = 26 \quad \Rightarrow \quad 3b + c = 8 \quad \text{(Equation 3)}
   $$

**Step 4: Solve the System of Equations**

Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 7 - 6 \quad \Rightarrow \quad b = 1
$$
Substitute $$ b = 1 $$ into Equation 1:
$$
1 + c = 6 \quad \Rightarrow \quad c = 5
$$

**Final Formula:**
$$
T_n = 2n^2 + n + 5
$$

**Verification:**
$$
\begin{align*}
n = 1: & \quad 2(1)^2 + 1 + 5 = 2 + 1 + 5 = 8 \
n = 2: & \quad 2(2)^2 + 2 + 5 = 8 + 2 + 5 = 15 \
n = 3: & \quad 2(3)^2 + 3 + 5 = 18 + 3 + 5 = 26 \
n = 4: & \quad 2(4)^2 + 4 + 5 = 32 + 4 + 5 = 41 \
n = 5: & \quad 2(5)^2 + 5 + 5 = 50 + 5 + 5 = 60 \
\end{align*}
$$
All terms match the given sequence.

---

**Summary of Answers:**

- **Problem 7:**
  - **(i)** $$ T_n = n^2 + 6 $$
  - **(ii)** $$ T_n = n^2 + 2n - 1 $$
  - **(iii)** $$ T_n = 2n^2 $$

- **Problem 8:**
  
  The $$ n $$th term of the sequence $$ 8,15,26,41,60,\ldots $$ is $$ T_n = 2n^2 + n + 5 $$.
**Problem 7 Solutions:** - **(i)** $$ T_n = n^2 + 6 $$ - **(ii)** $$ T_n = n^2 + 2n - 1 $$ - **(iii)** $$ T_n = 2n^2 $$ **Problem 8 Solution:** The $$ n $$th term is $$ T_n = 2n^2 + n + 5 $$.

Find an expression for the th term of each of the quadratic sequences:

Show that the th term of the quadratic sequence is given by

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Pre Algebra Questions

Find an expression for the th term of each of the quadratic sequences: Show that the th term of the quadratic sequence is given by

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Pre Algebra Questions

Find an expression for the th term of each of the quadratic sequences:

Show that the th term of the quadratic sequence is given by