An AP consists of 50 terms in which the term is 12 and the last term is 106. Find the term. (Hint: If is the first term and the common difference, then we arrive at the equations and . Solve this pair of linear equations for and .)

Understand the Question The term of an arithmetic progression (AP) is given by the formula: where is the first term and is the common difference. Given: Total number of terms term: Last term ( term): We form a system of two linear equations in and , solve for both, and then evaluate the term (). Step 1 · Find the First Term and Common Difference Using the term formula : For the term (): For the term (): Subtracting equation (1) from equation (2): Substituting into equation (1): Step 2 · Calculate the Term Using with , , and : Answer > 64 Common Mistakes Index error: Writing instead of . Sign errors during elimination: Subtracting the equations incorrectly (e.g. subtracting terms with wrong signs). Incorrect substitution: Substituting but forgetting to multiply by in .

Predicting What Comes Next: Sequences and Progressions | Exercise 8.2

Question 4

An AP consists of 50 terms in which the $3^{\text{rd}}$ term is 12 and the last term is 106. Find the $29^{\text{th}}$ term.

(Hint: If $a$ is the first term and $d$ the common difference, then we arrive at the equations $a + 2d = 12$ and $a + 49d = 106$ . Solve this pair of linear equations for $a$ and $d$ .)

Solution

Understand the Question

The $n^{\text{th}}$ term of an arithmetic progression (AP) is given by the formula: $a_n = a + (n - 1)d$ where $a$ is the first term and $d$ is the common difference.
Given:
- Total number of terms $= 50$
- $3^{\text{rd}}$ term: $a_3 = 12$
- Last term ( $50^{\text{th}}$ term): $a_{50} = 106$
We form a system of two linear equations in $a$ and $d$ , solve for both, and then evaluate the $29^{\text{th}}$ term ( $a_{29}$ ).

Step 1 · Find the First Term $a$ and Common Difference $d$

Diagram 1

Using the $n^{\text{th}}$ term formula $a_n = a + (n-1)d$ :

For the $3^{\text{rd}}$ term ( $a_3 = 12$ ):

\begin{aligned} a + (3-1)d &= 12 \\ a + 2d &= 12 \quad \dots (1) \end{aligned}

For the $50^{\text{th}}$ term ( $a_{50} = 106$ ):

\begin{aligned} a + (50-1)d &= 106 \\ a + 49d &= 106 \quad \dots (2) \end{aligned}

Subtracting equation (1) from equation (2):

\begin{aligned} (a + 49d) - (a + 2d) &= 106 - 12 \\ 47d &= 94 \\ d &= 2 \end{aligned}

Substituting $d = 2$ into equation (1):

\begin{aligned} a + 2(2) &= 12 \\ a + 4 &= 12 \\ a &= 12 - 4 \\ a &= 8 \end{aligned}

Step 2 · Calculate the $29^{\text{th}}$ Term

Using $a_n = a + (n-1)d$ with $a = 8$ , $d = 2$ , and $n = 29$ :

\begin{aligned} a_{29} &= a + (29-1)d \\ a_{29} &= 8 + (28)(2) \\ a_{29} &= 8 + 56 \\ a_{29} &= 64 \end{aligned}

Answer

64

Common Mistakes

Index error: Writing $a_{50} = a + 50d$ instead of $a + (50-1)d = a + 49d$ .
Sign errors during elimination: Subtracting the equations incorrectly (e.g. subtracting terms with wrong signs).
Incorrect substitution: Substituting $d = 2$ but forgetting to multiply by $2$ in $a + 2d = 12$ .

Question 4

More questions in Exercise 8.2