NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3
Question 22

Question. 22

Which term of the AP \(-2,-7,-12,\ldots\) is \(-77\)? Find the sum up to that term.

Answer:

16th term; sum = −632

Detailed Answer with Explanation:

Step 1: Identify the first term and common difference.

The AP is: \(-2, -7, -12, \ldots\)

So, first term \(a = -2\).

Common difference \(d = -7 - (-2) = -7 + 2 = -5\).

Step 2: Use the nth term formula of an AP.

The nth term is given by: \(a_n = a + (n-1)d\).

We are told that \(a_n = -77\).

So, \(-77 = -2 + (n-1)(-5)\).

Step 3: Solve for n.

\(-77 = -2 - 5(n-1)\)

\(-77 + 2 = -5(n-1)\)

\(-75 = -5(n-1)\)

Divide both sides by -5:

\(n-1 = 15\)

So, \(n = 16\).

Therefore, \(-77\) is the 16th term of the AP.

Step 4: Find the sum up to 16 terms.

Sum of first n terms is given by: \(S_n = \dfrac{n}{2}[a + a_n]\).

Here, \(n = 16, a = -2, a_{16} = -77\).

So, \(S_{16} = \dfrac{16}{2}[(-2) + (-77)]\).

\(S_{16} = 8[-79]\)

\(S_{16} = -632\).

Final Answer: The required term is the 16th term and the sum up to it is \(-632\).

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