Question

sum of three numbers of an ap is 16 and sum of their cubes is 792 find the numbers
​

Answer 1

Solution :

$\bigstar$Firstly, Let the three numbers of an A.P. are;

• (a-d)
• a
• (a+d)

$\underbrace{\bf{According\:to\:the\:question\::}}}}$

$\longrightarrow\sf{(a-d)+a+(a+d)=12}\\\\\longrightarrow\sf{a\cancel{-d}+a+a\cancel{+d}=12}\\\\\longrightarrow\sf{3a=12}\\\\\longrightarrow\sf{a=\cancel{12/3}}\\\\\longrightarrow\bf{a=4}$

&

$\longrightarrow\sf{(a-d)^{3} +a^{3} +(a+d)^{3} =792}\\\\\longrightarrow\sf{a^{3} \cancel{-d^{3}-3a^{2}} d+3ad^{2} +a^{3} +a^{3} \cancel{+d^{3} +3a^{2} d} +3ad^{2} =792\:\:\:[\therefore \:using\:(a-b)^{3} \:\:\& \:\:(a+b)^{3} ]}\\\\\longrightarrow\sf{3a^{3} +6ad^{2}=792}\\\\\longrightarrow\sf{3(4)^{3}+6(4)d^{2}=792\:\:\:[\therefore a=4]}\\\\\longrightarrow\sf{3\times 64+24d^{2}=792}\\\\\longrightarrow\sf{192+24d^{2}=792}\\\\\longrightarrow\sf{24d^{2}=792-192}\\\\\longrightarrow\sf{24d^{2}=600}$$\longrightarrow\sf{d^{2}=600/24}\\\\\longrightarrow\sf{d^{2}=25}\\\\\longrightarrow\sf{d=\pm\sqrt{25} }\\\\\longrightarrow\bf{d=\pm5}$

$\boxed{\bf{A\:R\:I\:T\:H\:M\:A\:T\:I\:C\:\:\:N\:U\:M\:B\:E\:R'S\::}}}}$

Using +ve common difference :

$\bullet\sf{(a-d)=4-5=\boxed{\bf{-1}}}\\\\\bullet\sf{a=\boxed{\bf{4}}}\\\\\bullet\sf{(a+d)=4+5=\boxed{\bf{9}}}$

Using -ve common difference :

$\bullet\sf{(a-d)=4-(-5)=4+5=\boxed{\bf{9}}}\\\\\bullet\sf{a=\boxed{\bf{4}}}\\\\\bullet\sf{(a+d)=4+(-5)=4-5=\boxed{\bf{-1}}}$