Question

Three positive integers a1 a2, a3 are in AP such that a1 + a2 + a3 = 33 and a1 x a2x a3 = 1155. Find the integers a1, a2 and a3. ans step by step pls no spam

Answer 1

$\text{Given:}$

$a_1,\;a_2,\;a_3,\;\text{are in A.P}$

$\implies\\\\a_1=a-d\\\\a_2=a\\\\a_3=a+d\text{ (say)}$

$a_1+a_2+a_3=33$

$(a-d)+a+(a+d)=33$

$3a=33$

$\implies\bf\,a=11$

$\text{Also }a_1{\times}a_2{\times}a_3=1155$

$\implies\,(a-d)a(a+d)=1155$

$\implies\,a(a^2-d^2)=1155$

$\implies\,11(11^2-d^2)=1155$

$\implies\,121-d^2=105$

$\implies\,121-105=d^2$

$\implies\,d^2=16$

$\implies\bf\,d={\pm}4$

$\text{when d=4}$

$a_1=a-d=7$

$a_2=a=11$

$a_3=a+d=15$

$\text{when d=-4}$

$a_1=a-d=15$

$a_2=a=11$

$a_3=a+d=7$

Answer 2

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