Determine the AP whose third term is 16 and whose term exceeds the term by 12.

Understand the Question The term of an Arithmetic Progression (AP) with first term and common difference is given by: We are given two conditions: Third term Seventh term exceeds fifth term by : We use these conditions to find the values of and , and then write the terms of the AP: Step 1 · Find the Common Difference Let the first term of the AP be and the common difference be . The term is given by: Given that the term exceeds the term by : Substitute the formula for and : Step 2 · Find the First Term Given that the third term is : Substitute into the equation: Step 3 · Form the Arithmetic Progression Using the first term and common difference , the consecutive terms are: Therefore, the required AP is Answer > Common Mistakes Sign Errors in Subtraction: Forgetting parentheses when evaluating , leading to incorrect sign distribution: , not or . Index Confusion: Writing instead of .

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Question 2

Determine the AP whose third term is 16 and whose $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by 12.

Solution

Understand the Question

The $n^{\text{th}}$ term of an Arithmetic Progression (AP) with first term $a$ and common difference $d$ is given by: $a_n = a + (n - 1)d$
We are given two conditions:
- Third term $a_3 = 16$
- Seventh term exceeds fifth term by $12$ : $a_7 - a_5 = 12$
We use these conditions to find the values of $a$ and $d$ , and then write the terms of the AP: $a, a+d, a+2d, a+3d, \dots$

Step 1 · Find the Common Difference $d$

Let the first term of the AP be $a$ and the common difference be $d$ .

The $n^{\text{th}}$ term is given by: $a_n = a + (n - 1)d$

Given that the $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by $12$ : $a_7 - a_5 = 12$

Substitute the formula for $a_7$ and $a_5$ :

\begin{aligned} [a + (7 - 1)d] - [a + (5 - 1)d] &= 12 \\[0.6em] (a + 6d) - (a + 4d) &= 12 \\[0.6em] 2d &= 12 \\[0.6em] d &= 6 \end{aligned}

Step 2 · Find the First Term $a$

Given that the third term is $16$ :

\begin{aligned} a_3 &= 16 \\[0.6em] a + 2d &= 16 \end{aligned}

Substitute $d = 6$ into the equation:

\begin{aligned} a + 2(6) &= 16 \\[0.6em] a + 12 &= 16 \\[0.6em] a &= 16 - 12 \\[0.6em] a &= 4 \end{aligned}

Step 3 · Form the Arithmetic Progression

Using the first term $a = 4$ and common difference $d = 6$ , the consecutive terms are:

\begin{aligned} a_1 &= a = 4 \\[0.6em] a_2 &= a + d = 4 + 6 = 10 \\[0.6em] a_3 &= a + 2d = 4 + 12 = 16 \\[0.6em] a_4 &= a + 3d = 4 + 18 = 22 \end{aligned}

Therefore, the required AP is $4, 10, 16, 22, \dots$

Answer

$4, 10, 16, 22, \dots$

Common Mistakes

Sign Errors in Subtraction: Forgetting parentheses when evaluating $a_7 - a_5$ , leading to incorrect sign distribution: $(a + 6d) - (a + 4d) = 2d$ , not $2a + 2d$ or $2d - 2a$ .
Index Confusion: Writing $a_3 = a + 3d$ instead of $a_3 = a + (3-1)d = a + 2d$ .

Question 2

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