NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3
Question 6

Question. 6

The 26th, 11th and last terms are \(0,\;3\) and \(-\dfrac{1}{5}\), respectively. Find the common difference and number of terms.

Answer:

\(d=-\dfrac{1}{5},\; n=27\)

Detailed Answer with Explanation:

Step 1: Write the formula for the \(n\)-th term of an AP.

In an arithmetic progression (AP), the \(n\)-th term is:

\(a_n = a + (n-1)d\)

where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

Step 2: Use the information about the 26th term.

The 26th term is given as 0.

So, \(a + 25d = 0\) … (1)

Step 3: Use the information about the 11th term.

The 11th term is given as 3.

So, \(a + 10d = 3\) … (2)

Step 4: Subtract the two equations to find \(d\).

(1) – (2):

\((a + 25d) - (a + 10d) = 0 - 3\)

\(15d = -3\)

\(d = -\dfrac{1}{5}\)

Step 5: Find the first term \(a\).

Put \(d = -\dfrac{1}{5}\) in equation (2):

\(a + 10(-\dfrac{1}{5}) = 3\)

\(a - 2 = 3\)

\(a = 5\)

Step 6: Use the last term to find the number of terms.

Last term is \(-\dfrac{1}{5}\).

So, \(a + (n-1)d = -\dfrac{1}{5}\)

\(5 + (n-1)(-\dfrac{1}{5}) = -\dfrac{1}{5}\)

Multiply through by 5 to clear the fraction:

\(25 - (n-1) = -1\)

\(25 + 1 = n - 1\)

\(n - 1 = 26\)

\(n = 27\)

Final Answer: Common difference \(d = -\dfrac{1}{5}\), Number of terms \(n = 27\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.3 | Detailed Answers