The first term of an AP is 5, the last term i | vedclass

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(A) Given that,First term $a = 5$Last term $l = 45$Sum of terms $S_{n} = 400$Using the formula for the sum of an $AP$: $S_{n} = \frac{n}{2}(a + l)$Sub

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(A) Given that,First term $a = 5$Last term $l = 45$Sum of terms $S_{n} = 400$Using the formula for the sum of an $AP$: $S_{n} = \frac{n}{2}(a + l)$Substituting the values: $400 = \frac{n}{2}(5 + 45)$$400 = \frac{n}{2}(50)$$400 = 25n$$n = \frac{400}{25} = 16$Now,using the formula for the $n^{th}$ term: $l = a + (n - 1)d$Substituting the values: $45 = 5 + (16 - 1)d$$45 - 5 = 15d$$40 = 15d$$d = \frac{40}{15} = \frac{8}{3}$Thus,the number of terms is $16$ and the common difference is $\frac{8}{3}$.

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.

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