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# Find the 20th term from the last for the AP: 3, 8, 13, …, 253.A) 100B) 125C) 148D) 158

Last updated date: 20th Jun 2024
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Hint: An arithmetic progression or AP is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
This fixed number is called the common difference of the AP.
Let us denote the first term of an AP by $\mathop a\nolimits_1$ , the second term by $\mathop a\nolimits_2$ ,... ., $\mathop n\nolimits^{th}$ term by $\mathop a\nolimits_n$, and the common difference by d.
Then the AP becomes $\mathop a\nolimits_{1,} {\text{ }}\mathop a\nolimits_{2,} {\text{ }}\mathop a\nolimits_{3,} {\text{ }}.....{\text{,}}\;\mathop a\nolimits_{n,}$
So, $\mathop a\nolimits_2 - \mathop a\nolimits_1 = \mathop a\nolimits_3 - \mathop a\nolimits_2 = ... = \mathop a\nolimits_n - \mathop a\nolimits_{n - 1} = d$.
Let the First term of AP = $\mathop a\nolimits_1 = a$
$\therefore$ the second term of AP = $\mathop a\nolimits_2 = \mathop a\nolimits_1 + d = a + d$
Hence, the third term of AP = $\mathop a\nolimits_3 = \mathop a\nolimits_2 + d = a + d + d$
$\mathop { \Rightarrow a}\nolimits_3 = a + 2d$
And the fourth term of AP = $\mathop a\nolimits_4 = \mathop a\nolimits_3 + d = a + 2d + d$
$\mathop { \Rightarrow a}\nolimits_4 = a + 3d$
So, the $\mathop n\nolimits^{th}$ term $\mathop a\nolimits_n$ of the AP with first term a and common difference d is given by
$\mathop { \Rightarrow a}\nolimits_n = a + \left( {n - 1} \right)d$
Use the above result to solve the given question.

Step 1: Given that:
The AP: 3, 8, 13, …, 253
Here, the first term, $\mathop a\nolimits_1 = a = 3$
Second term, $\mathop a\nolimits_2 = 8$
Thus, the common difference, $d = \mathop a\nolimits_2 - \mathop a\nolimits_1$
$\Rightarrow d = 8 - 3 = 5$
The last term of the given AP, $\mathop a\nolimits_n = 253$
Step 2: Find the total terms of the given AP.
To find the 11th term from the last term, we will find the total number of terms in the AP.
We know, $\mathop { \Rightarrow a}\nolimits_n = a + \left( {n - 1} \right)d$
Substituting the values from step 1
$\Rightarrow {\text{ }}253 = 3 + \left( {n - 1} \right)5 \\ \Rightarrow 253 - 3 = \left( {n - 1} \right)5 \\ \Rightarrow {\text{ }}250 = \left( {n - 1} \right)5 \\ \Rightarrow {\text{ }}\dfrac{{250}}{5} = \left( {n - 1} \right) \\ \Rightarrow {\text{ }}50 = n - 1 \\ \Rightarrow {\text{ }}n = 50 + 1 \\$
$\because$ $n = 51$
So, there are 51 terms in the given AP.
Step 3:
Let the 20th term from the last = p
We know,
The position of a term from the first = total numbers of the term – position of a term from the last + 1
Therefore, the position of term, p from the first = [51 – 20] + 1
$= 32$
Thus, the 20th term from the last term will be the 32nd term.
We know,
$\mathop { \Rightarrow a}\nolimits_n = a + \left( {n - 1} \right)d$
$\mathop { \Rightarrow {\text{ }}a}\nolimits_{32} = a + \left( {32 - 1} \right)d \\ \mathop { \Rightarrow {\text{ }}a}\nolimits_{32} = 3 + \left( {31} \right)5 \\ \mathop { \Rightarrow {\text{ }}a}\nolimits_{32} = 3 + 155 \\ \mathop { \Rightarrow {\text{ }}a}\nolimits_{32} = 158 \\$
The 20th term from the last term of the AP: 3, 8, 13, …, 253 is 158. Thus, the correct option is (D).

Note: Remember that the common difference of AP can be positive, negative, or zero.
Alternate method:
If we write the given AP in the reverse order, then $\mathop a\nolimits_1 = a = 253$
And $d = - 5$
So, the question now becomes finding the 20th term with these $a$ and $d$.
So,
$\mathop a\nolimits_{20} = a + \left( {20 - 1} \right)d$
$\mathop { \Rightarrow {\text{ }}a}\nolimits_{20} = 253 + \left( {19} \right)\left( { - 5} \right) \\ \mathop { \Rightarrow {\text{ }}a}\nolimits_{20} = 253 - 95 \\ \mathop { \Rightarrow {\text{ }}a}\nolimits_{20} = 158 \\$
So, the 20th term, which is now the required term, is 158.