Q.7. The first term of an A.P. of 30 non-negative terms is . If the sum of this A.P. is the cube of

Q: What is the correct answer for JEE Main Mathematics Q7?

The correct answer is A. Sol. Let the A.P. be $a_1, a_2, ..., a_{30}$. Given $a_1 = \frac{10}{3}$. Let $d$ be the common difference. The last term is $a_{30} = a_1 + (30-1)d = \frac{10}{3} + 29d$. The sum of the A.P. is $S_{30} = \frac{30}{2}(a_1 + a_{30}) = 15(\frac{10}{3} + \frac{10}{3} + 29d) = 15(\frac{20}{3} + 29d)$. We are given that $S_{30} = (a_{30})^3$. So, $15(\frac{20}{3} + 29d) = (\frac{10}{3} + 29d)^3$. Let $

Q: Q.7. The first term of an A.P. of 30 non-negative terms is . If the sum of this A.P. is the cube of its last term, then its common difference is :

The correct answer is A. Sol. Let the A.P. be $a_1, a_2, ..., a_{30}$. Given $a_1 = \frac{10}{3}$. Let $d$ be the common difference. The last term is $a_{30} = a_1 + (30-1)d = \frac{10}{3} + 29d$. The sum of the A.P. is $S_{30} = \frac{30}{2}(a_1 + a_{30}) = 15(\frac{10}{3} + \frac{10}{3} + 29d) = 15(\frac{20}{3} + 29d)$. We are given that $S_{30} = (a_{30})^3$. So, $15(\frac{20}{3} + 29d) = (\frac{10}{3} + 29d)^3$. Let $

Q: What are the options for this JEE Main Mathematics question?

A. \frac{5}{87} · B. \frac{25}{83} · C. \frac{15}{29} · D. \frac{5}{29}

Question

Q.7. The first term of an A.P. of 30 non-negative terms is . If the sum of this A.P. is the cube of its last term, then its common difference is :

FirstInTest · Accepted Answer

Sol. Let the A.P. be $a_1, a_2, ..., a_{30}$. Given $a_1 = \frac{10}{3}$. Let $d$ be the common difference. The last term is $a_{30} = a_1 + (30-1)d = \frac{10}{3} + 29d$. The sum of the A.P. is $S_{30} = \frac{30}{2}(a_1 + a_{30}) = 15(\frac{10}{3} + \frac{10}{3} + 29d) = 15(\frac{20}{3} + 29d)$. We are given that $S_{30} = (a_{30})^3$. So, $15(\frac{20}{3} + 29d) = (\frac{10}{3} + 29d)^3$. Let $X = \frac{10}{3} + 29d$. Then $15(\frac{10}{3} + X) = X^3$. This is $15(\frac{10}{3} + \frac{10}{3} + 29d) = (\frac{10}{3} + 29d)^3$. The solution provided in the image uses $S_{30} = 15(a_1 + a_{30}) = 15(\frac{10}{3} + \frac{10}{3} + 29d) = 15(\frac{20}{3} + 29d)$. It also states $a_{30} = \frac{10}{3} + 29d$. The equation is $15(\frac{20}{3} + 29d) = (\frac{10}{3} + 29d)^3$. Let $X = \frac{10}{3} + 29d$. Then $15(X + \frac{10}{3}) = X^3$. This is $15(X + a_1) = X^3$. The solution in the image has $S_{30} = (T_{30})^3$. $15(\frac{20}{3} + 29d) = (\frac{10}{3} + 29d)^3$. By observation, if $X = 5$, then $15(\frac{10}{3} + 5) = 15(\frac{25}{3}) = 125$. And $X^3 = 5^3 = 125$. So $X = 5$ is a solution. Thus, $\frac{10}{3} + 29d = 5$. $29d = 5 - \frac{10}{3} = \frac{15-10}{3} = \frac{5}{3}$. So $d = \frac{5}{3 	imes 29} = \frac{5}{87}$.

Answer

\frac{5}{87}

Answer

\frac{25}{83}

Answer

\frac{15}{29}

Answer

\frac{5}{29}

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