The first term of an A.P. of 30 non-negative | vedclass

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Question

(A) Let the first term be $a = \frac{10}{3}$ and the number of terms be $n = 30$.
The last term $L$ is given by $L = a + (n-1)d = \frac{10}{3} + 29d$

Vedclass · Accepted Answer

$\frac{5}{87}$

Vedclass · Answer

(A) Let the first term be $a = \frac{10}{3}$ and the number of terms be $n = 30$.
The last term $L$ is given by $L = a + (n-1)d = \frac{10}{3} + 29d$.
The sum of the $A$.$P$. is $S_{30} = \frac{n}{2}(a + L) = \frac{30}{2}(\frac{10}{3} + L) = 15(\frac{10}{3} + L) = 50 + 15L$.
Given that $S_{30} = L^3$, we have $L^3 = 15L + 50$, which implies $L^3 - 15L - 50 = 0$.
By testing values, if $L = 5$, then $5^3 - 15(5) - 50 = 125 - 75 - 50 = 0$.
Thus, $L = 5$ is a root.
Substituting $L = 5$ into the expression for the last term: $\frac{10}{3} + 29d = 5$.
$29d = 5 - \frac{10}{3} = \frac{15 - 10}{3} = \frac{5}{3}$.
Therefore, $d = \frac{5}{3 	imes 29} = \frac{5}{87}$.

The first term of an $A$.$P$. of $30$ non-negative terms is $\frac{10}{3}$. If the sum of this $A$.$P$. is the cube of its last term, then its common difference is:

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