Let $a,b,c$ be positive integers such that $b/a$ is an integer. If $a,b,c$ are in geometric progression and the arithmetic mean of $a,b,c$ is $b + 2$, then the value of $\frac{a^{2} + a - 14}{a + 1}$ is - Math Practice Question | divineJEE

Home
Test SeriesNEW!
About Us

Contact Us
Knowledge Check

Test Series

Mathematics
Sequence & Series

Mathematics

Select an option to instantly check whether it is correct or wrong.

Related Questions

Let the area of the region enclosed by the curves $y=3x,2y=27-3x$ and $y=3x-x \sqrt x$ be A. Then 10 A is equal to

If the area of the region $\begin{cases}(x,y):\frac{a}{x^2}≤ y ≤\frac{1}{x},1≤x≤2,0 < a < 1 \end{cases}$ is ${(log_e⁡2)}-\frac{1}{7}$ then the value of $7a-3$ is equal to

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2x$ is equal to:

The parabola $y^2=4x$ divides the area of the circle $x^2+y^2=5$ in two parts. The area of the smaller part is equal to:

The area (in square units) of the region enclosed by the ellipse $x^2+3y^2=18$ in the first quadrant below the line $y=x$ is

The area enclosed by the curves $xy+4y=16$ and $x+y=6$ is equal to :

The area (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2x-8$, is

The area of the region $\begin{cases}(x,y):y^2≤4x,x<4,\frac{xy(x-1)(x-2)}{(x-3)(x-4)}>0,x≠3\end{cases}$ is

The area of the region enclosed by the parabolas $y=4x-x^2$ and $3y=(x-4)^2$ is equal to

The area bounded by the curves $y=|x-1|+|x-2|$ and $y=3$ is equal to

Download Divine JEE App

Learn on the go with our mobile app. Access courses, live classes, and study materials anytime, anywhere.

Empowering students to achieve their IIT dreams through personalized learning and expert guidance.

Quick Links

About Us
Courses
Why Choose Us
Contact Us

Exam Categories

JEE Main
JEE Advanced
NEET
BITS Pilani

Contact Info

contact.divinejee@gmail.com
+91 9137402346
2nd floor, C, 10, Mahant Swami Marg, nr. Pratap Stadium, Chitrakoot, Jaipur, Rajasthan 302021