The number $3^{9} + 3^{12} + 3^{15} + 3^{n}$ is a perfect cube of an integer for natural number n, then n is:

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given expression is,

3^{9} + 3^{12} + 3^{15} + 3^{n}

Now, take

3^{9}

common from all the terms

{\Rightarrow 3}^{9} (1 + 3^{3} + 3^{6} + 3^{n - 9})

We know that,

(1 + x)^{3} = 1 + 3 x + x^{3} + 3 x^{2}

Put,

x = 3^{2}

\Rightarrow {(1 + 3^{2})}^{3} = 1 + 3 \times 3^{2} + {(3^{2})}^{3} + 3 \times {(3^{2})}^{2}

\Rightarrow {(1 + 3^{2})}^{3} = 1 + 3^{3} + 3^{6} + 3^{5}

The given expression should be compared to the above equation to get,

3^{5} = 3^{n - 9}

Compare the exponents of the same bas,

\Rightarrow n - 9 = 5

\Rightarrow n = 5 + 9

\Rightarrow n = 14

Hence, natural number n is 14.
Correct option is 1

Watch 3-min video & get full concept clarity