Question

find three consecutive terms in an AP whose sum is -3 and the product of their cubes is 512​

Answer 1

Solution :-

Let :-

First term, $\sf a_1 = a-d$

Second term, $\sf a_2 = a$

Third term, $\sf a_3 = a+d$

According to the first condition :-

$\longrightarrow \sf a_1+a_2+a_3 = -3 \\\\\longrightarrow a-d+a+a+d = -3 \\\\\longrightarrow 3a = -3 \\\\\longrightarrow \bf a = -1$

According to the second condition :-

$\longrightarrow \sf (a_1)^3\times (a_2)^3 \times (a_3)^3= 512 \\\\\longrightarrow (a-d)^3 \times a^3 \times(a+d)^3 = 512 \\\\\longrightarrow [ \ (a-d)(a)(a+d) \ ] ^3 = 512$

Substitute the value of a,

$\longrightarrow \sf [ \ (-1-d)(-1)(-1+d) \ ] ^3= 512 \\\\\longrightarrow [ \ -(d+1)(-1) (d-1) \ ] ^3= 512 \\\\\longrightarrow[ \ (d+1)(d-1) \ ]^3= 512 \\\\\longrightarrow d^2 - 1 = \sqrt[3]{512} \\\\\longrightarrow d^2 - 1 = 8 \\\\\longrightarrow d^2 = 9 \\\\\longrightarrow\bf d= \pm 3$

When d = 3,

$\bullet \sf \ a_1 = -1-3 = -4 \\\\\bullet \ a_2 = -1 \\\\\bullet \ a_3 = -1+3 = 2$

Three consecutive terms = -4 , -1 , 2

When d = -3,

$\bullet \ \sf a_1 = -1-(-3) = 2\\\\\bullet \ a_2 = -1 \\\\\bullet \ a_3 = -1-3 = -4$

Three consecutive terms = 2 , -1 , -4

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

Find three consecutive terms in an AP whose sum is -3 and the product of their cubes is 512

$\huge{\boxed{\rm{\red{Answer}}}}$

$\large{\boxed{\sf{Given \: that}}}$

• Sum of consecutive terms in AP is -3

• And the product of consecutive numbers cubes = 512

$\large{\boxed{\sf{To \: find}}}$

• Three consecutive terms of AP in given question.

$\large{\boxed{\sf{Assumptions}}}$

• First term = a¹ = a - d

• Second term = a² = a

• Third term = a³ = a + d

$\large{\boxed{\sf{Solution}}}$

• Three consecutive terms of AP = 2 , -1 and -4

$\large{\boxed{\sf{Full \: solution}}}$

$\large\pink{\texttt{According to 1st condition}}$

$\mapsto$ a¹ + a² + a³ = -3

$\mapsto$ a - d + a + a + d = -3

{ Actually the terms a - d + a + a + d in 2nd line of 1st condition are assumed by us }

$\mapsto$ 3a = -3

{ We write 3a bcoz there are 3 a's in our assumption ; Nd the 2nd thing we don't write d bcz d nd d cht each other }

$\mapsto$ a = - 1

{ Now we get a = -1..... thinking where does 3 of 3a nd 3 of -3 gone ?? It's simple guys :) 3 and 3 cut each other }

$\large\pink{\texttt{According to 1st condition}}$

$\mapsto$ (a1)³ × (a2)³ × (a3)³ = 512

$\mapsto$ (a-d)³ × a³ × (a+d)³ = 512

{ Actually the terms (a-d)³ × a³ × (a+d)³ in 2nd line of 2nd condition are assumed by us }

$\mapsto$ [ ( a-d )(a)(a+d) ]³ = 512

{ We combined power in this assumption bcz the power is same . And according to power nd exponential formula a² + b² = (ab)² }

$\large\purple{\texttt{Substituting the value of a we get following results}}$

$\mapsto$ [ ( -1-d )(-1)(-1+d) ]³ = 512

{ We put - 1 at the place of a bcz In condition 1 we get a = -1 that's whtmy we substitute -1 at the place of a }

$\mapsto$ [ - (d+1) (-1) (d-1) ]³ = 512

{In this We reverse the values }

$\mapsto$ [ ( d+1) (d-1) ]³ = 512

{ We get [ ( d+1) (d-1) ] from [ - (d+1) (-1) (d-1) ]³ } { Because - - be + so 1 and - is cutted }

$\mapsto$ d² - 1 = ³√513

{ We write d² instead in the place of ( d+1) (d-1) bcz we have two d's that why we write d² and we remove ³ from [ ( d+1) (d-1) ]³ and added it to 513 hence it becomes cube root 513 now }

$\mapsto$ d² - 1 = 8

{ We get 8 bcz cube root of 513 is always 8 }

$\mapsto$ d² = 8 + 1

{ We know that - = + }

$\mapsto$ d² = 9

{ By adding 8 + 1 we get 9 Easy !! }

$\mapsto$ d = √9

{ We get square root 9 coz ² = √ }

{ Hence d² = 9 is converted to d = √9 }

$\mapsto$ d = 3

{We get 3 coz square root 9 is always 3 }

$\large\pink{\texttt{According to 3rd condition}}$

$\large\pink{\texttt{When d = 3}}$

$\mapsto$ a1 = -1 -3 = (-4)

$\mapsto$ a2 = -1

$\mapsto$ a3 = -1 + 3 = -2

⚪Three consecutive terms are -4 , 2

nd -1 .

$\large\pink{\texttt{When d = -3}}$

☆ a1 = -1 - (-3) = 2

☆ a2 = -1

☆ a3 = - 1 - 3 = -4

$\large{\boxed{\boxed{\underbrace{\sf{3 \: consecutive \: terms \: are}}}}}$

$\large{\boxed{\boxed{\underbrace{\sf{2 , \: -1 , \: -4}}}}}$

Hope it's helpful

Thank you :)