NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3

Question 11

Key facts:

• In any sequence, the nth term can be found from sums using: \(a_n = S_n - S_{n-1}\).
• For an AP, the terms differ by a constant common difference \(d\).

Step 1: Use the given partial sums to find the first few terms

We know:

\[S_1 = a_1 = 3\]

So, \(a_1 = 3\).

Also,

\[S_2 = a_1 + a_2 = 4\]

So,

\[a_2 = S_2 - S_1 = 4 - 3 = 1\]

Hence, \(a_1 = 3\), \(a_2 = 1\).

The common difference is:

\[d = a_2 - a_1 = 1 - 3 = -2\]

Step 2: Write the general term \(a_n\)

For an AP,

\[a_n = a_1 + (n - 1)d\]

Substitute \(a_1 = 3\) and \(d = -2\):

\[a_n = 3 + (n - 1)(-2) = 3 - 2n + 2 = 5 - 2n\]

So,

\[a_n = 5 - 2n\]

This also matches the required form \(a_n = S_n - S_{n-1}\).

Step 3: Find \(S_3\) and \(a_3\)

First find \(a_3\) using \(a_n = 5 - 2n\):

\[a_3 = 5 - 2 \cdot 3 = 5 - 6 = -1\]

Now,

\[S_3 = a_1 + a_2 + a_3 = 3 + 1 + (-1) = 3\]

Check with difference:

\[a_3 = S_3 - S_2 = 3 - 4 = -1\]

which agrees with our value.

Step 4: Find \(a_{10}\)

Using the general term:

\[a_{10} = 5 - 2 \cdot 10 = 5 - 20 = -15\]

By definition, \(a_{10} = S_{10} - S_9\), so:

\[S_{10} - S_9 = -15\]

Step 5: Summary of all required results

• \(a_2 = S_2 - S_1 = 1\)
• \(S_3 = 3\)
• \(a_3 = S_3 - S_2 = -1\)
• \(a_{10} = S_{10} - S_9 = -15\)
• General term: \(a_n = S_n - S_{n-1} = 5 - 2n\).