Question:
If \(3x^2-2ax+(a^2+2b^2+2c^2) = 2(ab+bc),\) then a,b,c can be in
If \(3x^2-2ax+(a^2+2b^2+2c^2) = 2(ab+bc),\) then a,b,c can be in
Updated On: Sep 18, 2024
A.P
G.P
H.P
none of these
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
The given result can be expressed as follows :
\(\left\{x-(a-b)^2\right\}+\left\{x-(b-c)^2\right\}+(x-c)^2=0\)
\(x=a-b,x=b-c,x=c\)
\(a-b=b-c=c\)
From \(a-b=b-c,\ 2b=a+c\)
Therefore, \(a,b,c\) can be in A.P.
So, the correct option is (A) : A.P.
Was this answer helpful?
0
7
Top Questions on Arithmetic Progression
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to
- JEE Main - 2026
- Mathematics
- Arithmetic Progression
- Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to
- JEE Main - 2026
- Mathematics
- Arithmetic Progression
- The vertices B and C of a triangle ABC lie on the line \( \frac{x}{1} = \frac{1-y}{2} = \frac{z-2}{3} \). The coordinates of A and B are (1, 6, 3) and (4, 9, 6) respectively and C is at a distance of 10 units from B. The area (in sq. units) of \( \triangle ABC \) is:
- JEE Main - 2026
- Mathematics
- Arithmetic Progression
- The common difference of the A.P.: \( a_1, a_2, \ldots, a_m \) is 13 more than the common difference of the A.P.: \( b_1, b_2, \ldots, b_n \). If \( b_{31} = -277 \), \( b_{43} = -385 \) and \( a_{78} = 327 \), then \( a_1 \) is equal to:
- JEE Main - 2026
- Mathematics
- Arithmetic Progression
- If the sum of the first four terms of an A.P. is \(6\) and the sum of its first six terms is \(4\), then the sum of its first twelve terms is
- JEE Main - 2026
- Mathematics
- Arithmetic Progression
View More Questions