Question

If $a_1$, $a_2$, $a_3$, $a_4$ and $b$ are real numbers such that $(a_1^2+a_2^2+a_3^2)b^2-2(a_1{a}_2+a_2{a}_3+a_3{a}_4)b+(a_2^2+a_3^2+a_4^2)\le 0$ then $a_1$, $a_2$, $a_3$, $a_4$ are the terms of a/an

• A. G.P.
• B. H.P.
• C. A.G.P.
• D. A.P.

Answer 1

Answer: (A)

G.P.

Solution:

$(a_1^2+a_2^2+a_3^3)b^2-2(a_1{a}_2+a_2{a}_3+a_3{a}_4)b+(a_2^2+a_3^3+a_4^2)\le 0$

where, $a_1,a_2,a_3,a_4,b\in R$

or, $(a_1^2{b}^2-2a_1{a}_{2}b+a_2^2)+(a_2^2{b}^2-2a_2{a}_{3}b+a_3^2)+(a_3^2{b}^2-2a_3{a}_{4}b+a_4^2)\le 0$

or, $(a_{1}b-a_2{)}^2+(a_{2}b-a_3{)}^2+(a_{3}b-a_4{)}^2\le 0$

Since all individual terms are positive so, they all are equal to zero separately

or, $(a_{1}b-a_2{)}^2=0$ or $(a_{2}b-a_3{)}^2=0$or $(a_{3}b-a_4{)}^2=0$

or, $(a_{1}b-a_2)=0$ or $(a_{2}b-a_3)=0$or $(a_{3}b-a_4)=0$

or, $b=\frac{a_2}{a_1}$ or $b=\frac{a_3}{a_2}$ or $b=\frac{a_4}{a_3}$

or, $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=b$

So, $a_1,a_2,a_3$ and $a_4$ are in G.P.

Hence, A is the correct option.