Question

What value of k will k+9, 2k-1 and 2k+7 are the consecutive terms of an AP?

Answer 1

Answer:

k = 18

Explanation:

Let a1 = k+9, a2=2k-1, and

a3 = 2k+7 are in A.P

We know that ,

a2 - a1 = a3 - a2

=> (2k-1)-(k+9) = (2k+7)-(2k-1)

=> 2k-1-k-9 = 2k+7-2k+1

= k-10 = 8

=> k = 8+10

=> k = 18

Therefore,

k = 18

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Answer 2

Answer: k = 18

Step-by-step explanation:

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let there will be three consecutive terms $a_1 , a_2 , a_3$ in an ap.

$a_1 = k+9\\\\a_2= 2k-1\\\\a_3 = 2k+7$

now as we know ,

$a_2 - a_1 = a_3 - a_2$ = common difference.

on putting the values ,

(2k-1) -( k+9) = (2k+7)- (2k-1)

=> 2k - 1 - k -9 = 2k + 7 - 2k + 1

=> k - 10 = 8

=> k = 8 + 10

=> k= 18

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hope it helps :)

thanks :)