Question

determine






K so that K + 2, 4 k - 6, 3 k - 2 are the three consecutive terms of an AP

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Answer 1

$\huge \mathfrak{hello \: mate}$

We know that,

a2-a1=a3-a2

So,

4k-6-(k+2)=3k-2-(4k-6)

4k-6-k-2=3k-2-4k+6

3k-8=4-k

4k=12

k=3

Answer 2

$\huge\bf\pink{\mid{\overline{\underline{Your\: Answer}}}\mid}$

K = 3

step-by-step explanation:

Let the given terms be,

a = k +2

b = 4k -6

c = 3k -2

Now,

we know that,

if three consecutive terms a, b and c are in A.P,

then,

a + c = 2a

=> k+2 + 3k - 2 = 2( 4k -6)

on simplifying,

we get,

=> 4k = 8k - 12

=> 8k - 4k = 12

=> 4k = 12

=> k = 12/4

=> k = 3

Hence,

the value of k so that the terms are kn A.P is 3.