NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.1
Question 13

Question.  13

The 4th term from the end of the AP: \(-11,-8,-5,\ldots,49\) is

(A)

37

(B)

40

(C)

43

(D)

58

Detailed Answer with Explanation:

Step 1: Identify the first term and common difference.

The first term \(a = -11\).

The numbers are increasing by 3 each time, so the common difference \(d = 3\).

Step 2: Find how many terms are there in the AP.

The last term is given as \(l = 49\).

The formula to find the number of terms in an AP is:

\(n = \dfrac{l - a}{d} + 1\).

Substitute the values: \(n = \dfrac{49 - (-11)}{3} + 1 = \dfrac{60}{3} + 1 = 20 + 1 = 21\).

So, there are 21 terms in this AP.

Step 3: Understand what "4th term from the end" means.

If there are 21 terms in total, the last term is the 21st term. The 4th from the end means:

21st → last term

20th → 2nd from the end

19th → 3rd from the end

18th → 4th from the end

So we need to find the 18th term.

Step 4: Find the 18th term.

The formula for the \(n\)th term is: \(a_n = a + (n-1)d\).

Put \(n = 18\):

\(a_{18} = -11 + (18 - 1) \times 3\).

\(a_{18} = -11 + 17 \times 3\).

\(a_{18} = -11 + 51 = 40\).

Final Answer: The 4th term from the end is 40.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.1 | Detailed Answers