if 3x+K,2x+9and x+13 are three consecutive terms of an AP, find the value of K .​

1 thought on “if 3x+K,2x+9and x+13 are three consecutive terms of an AP, find the value of K .​”

1. Step-by-step explanation:

   Given :–

   3x+K,2x+9and x+13 are three consecutive terms of an AP

   To find :–

   Find the value of K ?

   Solution:–

   Method –1:–

   Given that :

   3x+K,2x+9and x+13 are three consecutive terms of an AP

   We know that

   In an AP , The Common difference is same throughout the series

   => Common difference = tn – tn-1

   => (2x+9)-(3x+K) = (x+13)-(2x+9)

   => 2x+9-3x-K = x+13-2x-9

   => 9-x-K = 4-x

   => 9-x-K-4+x = 0

   => 5-K = 0

   => K = 5

   Therefore, K = 5

   Method-2:–

   We know

   a,b,c are three consecutive terms in an AP then

   b = (a+c)/2

   We have

   a = 3x+K

   b = 2x+9

   c = x+13

   2x+9 = (3x+K+x+13)/2

   => 2x+9 = (4x+13+K)/2

   => 2(2x+9) = (4x+13+K)

   => 4x+18 = 4x+13+K

   => 4x+18-4x-13 = K

   => (4x-4x)+(18-13) = K

   => 0+5 = K

   => K = 5

   Therefore, K = 5

   Answer:–

   The value of K for the given problem is 5

   Used formulae:–

   • Common difference = tn – tn-1

   • tn = nth term

   • tn-1 = (n-1)th term

   • a,b,c are three consecutive terms in an AP then b = (a+c)/2

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