The sum of three consecutive terms of an A.P. | vedclass

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Question

(C) Let the three consecutive terms of the $A.P.$ be $(a-d)$,$a$,and $(a+d)$.Given that the sum of these terms is $30$:$(a-d) + a + (a+d) = 30$$3a = 3

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let the three consecutive terms of the $A.P.$ be $(a-d)$,$a$,and $(a+d)$.Given that the sum of these terms is $30$:$(a-d) + a + (a+d) = 30$$3a = 30$$a = 10$Given that the product of the first and last term is $75$:$(a-d)(a+d) = 75$$a^2 - d^2 = 75$Substituting $a = 10$:$10^2 - d^2 = 75$$100 - d^2 = 75$$d^2 = 100 - 75 = 25$$d = \pm 5$Since $5$ is given as an option,the common difference is $5$.

The sum of three consecutive terms of an $A.P.$ is $30$ and the product of its first and last term is $75$. Then,the common difference of the $A.P.$ is $\ldots \ldots \ldots \ldots .$

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