Question

Find k so that k, k+2, k+6 are consecutive terms of G.P​

Answer 1

Answer:

k=2

Step-by-step explanation:

Given terms are in G.P.

So that k+6/k+2=k+2/k

by cross multiplication we get,

k^2+6k=k^2+2k+2k+4

k^2+6k=K^2+4k+4

k^2+6k-k^2-4k-4=0

2k-4=0

2k=4

k=2

Answer 2

$\sf \huge \purple{ \underline{ \fbox{ \: \red{Solution \: \: : }}}}$

Given ,

k , k + 2 , k + 6 are forms an GP

Therefore ,

$\hookrightarrow \sf \frac{k + 2}{k} = \frac{k + 6}{k + 2} \\ \\ \sf By \: cross \: multiplication \: , \: we \: obtain \\ \\ \hookrightarrow \sf {( k + 2)}^{2} = {k}^{2} + 6k \\ \\ \hookrightarrow \sf {k}^{2} + {(2)}^{2} + 2(k)(2) = {k}^{2} + 6k \\ \\ \hookrightarrow \sf {k}^{2} + 4 + 4k = {k}^{2} + 6k \\ \\ \hookrightarrow \sf 2k = 4 \\ \\ \hookrightarrow \sf k = 2$

Hence , the value of k is 2