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

Question 5

Given: First term \(a = 5\), last term \(l = 45\), sum of all terms \(S_n = 400\).

We need to find the number of terms \(n\) and the common difference \(d\).

Step 1: Use the sum formula with first and last term

For an AP, the sum of \(n\) terms is:

\[S_n = \dfrac{n}{2}(a + l)\]

Substitute the given values:

\[400 = \dfrac{n}{2}(5 + 45) = \dfrac{n}{2} \cdot 50\]

So,

\[400 = 25n\]

\[n = \dfrac{400}{25} = 16\]

Number of terms: \(n = 16\).

Step 2: Use the nth term formula to find \(d\)

The last term \(l\) is the \(n\)th term:

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

Here, \(a_n = l = 45\), \(a = 5\), \(n = 16\).

So,

\[45 = 5 + (16 - 1)d\]

\[45 = 5 + 15d\]

Subtract 5 from both sides:

\[40 = 15d\]

\[d = \dfrac{40}{15} = \dfrac{8}{3}\]

Common difference: \(d = \dfrac{8}{3}\).

Conclusion: The AP has 16 terms and common difference \(\dfrac{8}{3}\).