Question

sum of 3 numbers of a.p. is 27 and sum of their squares is 261 ;find the numbers

Answer 1

Answer:

let the three numbers in A.P. be x, x-1,x+1

sum of 3 numbers of a.p. is 27

x+x-1+x+1=27

3x=27

x=27/3

x=9

so the numbers are 9,9+1=10,9-1=8.

Step-by-step explanation:

Answer 2

Solution :

$\bigstar$Let the three number's of an A.P. are;

• a-d
• a
• a+d

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

$\longrightarrow\sf{a\cancel{-d}+a+a\cancel{+d}=27}\\\\\longrightarrow\sf{3a=27}\\\\\longrightarrow\sf{a=\cancel{27/3}}\\\\\longrightarrow\bf{a=9}$

&

$\longrightarrow\sf{(a-d)^{2} +a^{2} +(a+d)^{2} =261}\\\\\longrightarrow\sf{a^{2}+d^{2} -2ad+a^{2} +a^{2} +d^{2} +2ad=261\:\:\underbrace{\rm{using\:formula\:of\:(a+b)^{2} }}}\\\\\longrightarrow\sf{3a^{2} +2d^{2} \cancel{-2ad+2ad} =261}\\\\\longrightarrow\sf{3a^{2} +2d^{2} =261}\\\\\longrightarrow\sf{3(9)^{2} +2d^{2} =261\:\:[\therefore a=9]}\\\\\longrightarrow\sf{3\times 81+2d^{2} =261}\\\\\longrightarrow\sf{243+2d^{2} =261}\\\\\longrightarrow\sf{2d^{2} =261-243}$

$\longrightarrow\sf{2d^{2} =18}\\\\\longrightarrow\sf{d^{2} =\cancel{18/2}}\\\\\longrightarrow\sf{d^{2} =9}\\\\\longrightarrow\sf{d=\pm\sqrt{9} }\\\\\longrightarrow\bf{d=\pm\:3}$

Now;

$\underline{\boldsymbol{The\:Number's\:of\:Arithmetic\:progression\::}}}}}$

Using for +ve common difference (d) :

$\bullet\sf{a-d=9-3=\boxed{\bf{6}}}\\\\\bullet\sf{a=\boxed{\bf{9}}}\\\\\bullet\sf{a+d=9+3=\boxed{\bf{12}}}$

Using for -ve common difference (d) :

$\bullet\sf{a-d=9-(-3)=9+3=\boxed{\bf{12}}}\\\\\bullet\sf{a=\boxed{\bf{9}}}\\\\\bullet\sf{a+d=9+(-3)=9-3=\boxed{\bf{6}}}$