If 3+k,18-k,5k+1 are in A. P, then find k

• 3 + k , 18 - k and 5k + 1 are in AP .

• The value of K .

• As we know that, In AP common difference is same :

• b - a = c - b

=> 2b = a + c

Here ,

• a = 3 + k

• b = 18 - k

• c = 5k + 1

=> 2 ( 18 - k ) = 3 + k + 5k + 1.

=> 36 - 2k = 4 + 6k

=> 36 - 4 = 6k + 2k

=> 32 = 8k

=> k = 32/8

=> k = 4

Hence , the value of k is 4.

k=4

Step-by-step explanation:

Since the given terms are in A.P., the common difference will be the same

So, 18-k-3-k = 5k+1-18+k

15-2k = 6k-17

6k+2k = 17+15

8k = 32

k = 4

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

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step-by-step explanation: