The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
\(n = 16,\ d = \dfrac{8}{3}\)
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 \(d\).
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: \(n = 16\).
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 \(\dfrac{8}{3}\).