NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3
Question 25

Question. 25

In an AP, if \(S_n=3n^2+5n\) and \(a_k=164\), find \(k\).

Answer:

\(k=27\)

Detailed Answer with Explanation:

Step 1: We know that \(S_n\) is the sum of the first \(n\) terms of the AP.

Step 2: The formula to find the \(n^{th}\) term is:

\(a_n = S_n - S_{n-1}\)

This is because the difference between the sum of first \(n\) terms and the sum of first \((n-1)\) terms gives the \(n^{th}\) term.

Step 3: Write \(S_n = 3n^2 + 5n\).

Now find \(S_{n-1}\):

\(S_{n-1} = 3(n-1)^2 + 5(n-1)\)

\(= 3(n^2 - 2n + 1) + 5n - 5\)

\(= 3n^2 - 6n + 3 + 5n - 5\)

\(= 3n^2 - n - 2\)

Step 4: Now find \(a_n = S_n - S_{n-1}\):

\(a_n = (3n^2 + 5n) - (3n^2 - n - 2)\)

\(= 3n^2 + 5n - 3n^2 + n + 2\)

\(= 6n + 2\)

Step 5: The general term is \(a_n = 6n + 2\).

We are told \(a_k = 164\).

So, \(6k + 2 = 164\).

Step 6: Solve for \(k\):

\(6k = 164 - 2 = 162\)

\(k = 162 / 6 = 27\)

Final Answer: \(k = 27\).

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