Answer

Verified

388.9k+ views

**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.

**Complete step-by-step answer:**

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 \\

$

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 { \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 \\

$

$

\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.**Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

The states of India which do not have an International class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you graph the function fx 4x class 9 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Name the three parallel ranges of the Himalayas Describe class 9 social science CBSE