Question

for what value of k will k+8, 3k-2 and 4k-7 are in AP​

Answer 1

Answer:

Given that k  

2

+4k+8,2k  

2

+3k+6,3k  

2

+4k+4 are in A.P.

Therefore, the difference between any two consecutive numbers is same.

Hence, (2k  

2

+3k+6)−(k  

2

+4k+8)=(3k  

2

+4k+4)−(2k  

2

+3k+6)

⟹k  

2

−k−2=k  

2

+k−2

⟹k+k=0

⟹2k=0

⟹k=0

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Step-by-step explanation: