The value of $x$ for which the numbers $2 x + 1, 5 x + 1$ and $11 x + 1$ are in G.P. is ___ ( $x$ is a natural number).

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 1
If $a, b, c$ are in G.P, then $b^{2} = a c .$

Hence, for $2 x + 1, 5 x + 1, a n d 11 x + 1$ to form a G.P., we must have

$(5 x + 1)^{2} = (2 x + 1) (11 x + 1) .$

$\Rightarrow 25 x^{2} + 10 x + 1 = 22 x^{2} + 13 x + 1$

$\Rightarrow 3 x^{2} - 3 x = 0$

$\Rightarrow 3 x (x - 1) = 0$

$\Rightarrow x = 0$ or $x = 1$

Since $x$ is a natural number, $2 x + 1, 5 x + 1$ and $11 x + 1$ are in G.P when $x = 1.$

Suggest Corrections

Similar questions

The value of x for which the numbers 2x+1, 5x+1, and 11x+1 are in G.P is ___ (x is a natural number).

f (x) = 2 x + 1

x

f (x), f (2 x), f (4 x)

Let f(x) = 2x + 1. Then the number of real values of x for which the three numbers f(x), f(2x), f(4x) are in G.P. is

^{'} a^{'}

f (x) = a - 1 + 2 x - x^{2}

g (x) = x^{2} - 2 a x + 10 - 2 a

f (x) = 2 x + 1

x

f (x), f (2 x), f (4 x)

G P

Join BYJU'S Learning Program