Question

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

Answer 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