Which term of the AP : will be more than its term?

Understand the Question In an Arithmetic Progression (AP), the term is given by the formula: where is the first term and is the common difference. We need to find the value of the term (), add to it to get the required term's value, and then determine the term number that corresponds to this value. Step 1 · Find First Term and Common Difference For the given AP: Step 2 · Calculate the 54th Term Using the term formula Step 3 · Find Value of the Required Term The required term is more than Step 4 · Determine the Term Number Let the required term be the term, so Answer > term Common Mistakes Adding to Term Index Instead of Term Value: Adding to the term index () rather than to the actual value of . Off-by-One Error: Forgetting to add in the final step after finding . Direct Shortcut: Since , you can also directly write .

Arithmetic Progressions | Exercise 5.2

Question 11

Which term of the AP : $3, 15, 27, 39, \dots$ will be $132$ more than its $54^{\text{th}}$ term?

Solution

Understand the Question

In an Arithmetic Progression (AP), the $n^{\text{th}}$ term is given by the formula: $a_n = a + (n - 1)d$ where $a$ is the first term and $d$ is the common difference.
We need to find the value of the $54^{\text{th}}$ term ( $a_{54}$ ), add $132$ to it to get the required term's value, and then determine the term number $n$ that corresponds to this value.

Step 1 · Find First Term and Common Difference

For the given AP: $3, 15, 27, 39, \dots$ Diagram 1

\begin{aligned} a &= 3 \\[0.6em] d &= 15 - 3 \\[0.6em] &= 12 \end{aligned}

Step 2 · Calculate the 54th Term

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

\begin{aligned} a_{54} &= 3 + (54-1) \times 12 \\[0.6em] &= 3 + 53 \times 12 \\[0.6em] &= 3 + 636 \\[0.6em] &= 639 \end{aligned}

Step 3 · Find Value of the Required Term

The required term is $132$ more than $a_{54}$

\begin{aligned} \text{Required term value} &= a_{54} + 132 \\[0.6em] &= 639 + 132 \\[0.6em] &= 771 \end{aligned}

Step 4 · Determine the Term Number $n$

Let the required term be the $n^{\text{th}}$ term, so $a_n = 771$

\begin{aligned} 771 &= 3 + (n-1) \times 12 \\[0.6em] 771 - 3 &= (n-1) \times 12 \\[0.6em] 768 &= (n-1) \times 12 \\[0.6em] \dfrac{768}{12} &= n - 1 \\[0.6em] 64 &= n - 1 \\[0.6em] n &= 64 + 1 \\[0.6em] n &= 65 \end{aligned}

Answer

$65^{\text{th}}$ term

Common Mistakes

Adding to Term Index Instead of Term Value: Adding $132$ to the term index ( $54 + 132 = 186$ ) rather than to the actual value of $a_{54}$ .
Off-by-One Error: Forgetting to add $1$ in the final step after finding $n - 1 = 64$ .
Direct Shortcut: Since $a_n - a_{54} = (n - 54)d$ , you can also directly write $(n - 54) \times 12 = 132 \implies n - 54 = 11 \implies n = 65$ .

Question 11

More questions in Exercise 5.2