Question

"Question3
Aim-1 , Progressions ,Challengers .Page - 116
find value of k such that k+2 , 4k-6 , 3k-2 are in A. P "

Answer 1

Given,
k + 2 , 4k - 6 , 3k - 2 are in A. P

We know that, Common difference is difference between any two consecutive terms.
So,
(4k - 6 )-(k+2 ) = d
(3k - 2 ) - ( 4k-6) = d

We know that, d = d
→ 4k - 6 - ( k + 2 ) = 3k - 2 - ( 4k - 6 )
→ 4k - 6 -k - 2 = 3k - 2 -4k + 6
→ 4k - k -8 = -k + 4
→ 3k - 8 = 4 - k
→ 3k + k = 4 + 8
→ 4k = 12
→ k = 12/4 = 3 .

Therefore, For k+2 , 4k - 6 , 3k - 2 to be in A. P, The value of K should be 3 .

Answer 2

Hey ❗❗

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↪To be terms of an AP the difference between two consecutive terms must be same

↪So according to the question k+2,4k-6 & 3k-2 are the terms of an AP

↪Now, 4k - 6 - (k+2) = 3k -2 - (4k-6)

↪4k -6 -k- 2 = 3k -2 - 4k+ 6

↪4k -k -6- 2 = 3k- 4k- 2+ 6

↪3k - 8 = -k + 4

↪3k +k -8- 4 = 0

↪4k - 12 = 0

↪4k = +12

↪k = 12/4

↪k = 3

Hence value of k is 3

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