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: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(1)

We have, $a = \frac{10}{3}$

Then, $l = \frac{10}{3} + 29 d$

Given, $S = l^{3}$

$\Rightarrow \frac{30}{2} [\frac{20}{3} + 29 d] = {(\frac{10}{3} + 29 d)}^{3}$

$\Rightarrow 15 [\frac{20}{3} + 29 d] = {(\frac{10}{3} + 29 d)}^{3}$

From options, $d = \frac{5}{87}$

Solution

Q.
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: [2026]

1 $\frac{5}{87}$

2 $\frac{25}{83}$

3 $\frac{15}{29}$

4 $\frac{5}{29}$

Ans.
(1)

We have, $a = \frac{10}{3}$

Then, $l = \frac{10}{3} + 29 d$

Given, $S = l^{3}$

$\Rightarrow \frac{30}{2} [\frac{20}{3} + 29 d] = {(\frac{10}{3} + 29 d)}^{3}$

$\Rightarrow 15 [\frac{20}{3} + 29 d] = {(\frac{10}{3} + 29 d)}^{3}$

From options, $d = \frac{5}{87}$

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Solution

Q. The first term of an A.P. of 30 non-negative terms is 103. If the sum of this A.P. is the cube of its last term, then its common difference is: [2026]

1 587

2 2583

3 1529

4 529

Ans. (1) We have, a=103 Then, l=103+29d Given, S=l3 ⇒ 302[203+29d]=(103+29d)3 ⇒ 15[203+29d]=(103+29d)3 From options, d=587