Find the values of k for which
Question:

Find the values of k for which k + 12, k – 6 and 3 are in GP.

Solution:

To find: Value of k

Given: k + 12, k – 6 and 3 are in GP

Formula used: (i) when $a, b, c$ are in GP $b^{2}=a c$

As, $k+12, k-6$ and 3 are in GP

$\Rightarrow(k-6)^{2}=(k+12)(3)$

$\Rightarrow \mathrm{k}^{2}-12 \mathrm{k}+36=3 \mathrm{k}+36$

$\Rightarrow \mathrm{k}^{2}-15 \mathrm{k}=0$

$\Rightarrow \mathrm{k}(\mathrm{k}-15)=0$

$\Rightarrow \mathrm{k}=0$, Or $\mathrm{k}=15$

Ans) We have two values of k as 0 or 15 

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