Question

There are 4 positive numbers k, k^2, k^3, k^4. If we subtract 12 by first number and 36 by last number, we will get an AP of 4 numbers. Calculate k?

Answer 1

Answer:

hello bhai

Step-by-step explanation:

as you told in your question in 4 positive number k , k^2 , k^3 , k^4 when we subtract 12 by first number so then the first number will be k-12

and 36 by last number so the last number will be k^4-36

so we can say that first term of a.p = a1 = k-12

a2=k^2

a3= k^3

but

a4= k^4-36

we have learnt that in the A.P

fourth term - third term = third term - second term                                 .........(1)

third term - second term =second term - first term                                ..........(2)

(its because the common difference d is always same any term of a.p )

now on these base we will make equation 1 and 2

  in equation (1)-:

fourth term - third term = third term - second term

or a4 - a3 = a3 - a2

or k^4-36 -  k^3 =  k^3 - k^2

after taking common on both side we can say that-

k^3(k - 1) - 36 = k^2(k - 1)

after changing the side of 36

k^3(k - 1) -  k^2(k - 1) = 36

(k^3 - k^ 2)( k - 1) = 36

[k^2(k-1)](k-1) = 36

k^2(k-1)(k-1) = 36                                                                    ..............(3)

from second equation

third term - second term =second term - first term                                ..........(2)

or a3 - a2 = a2- a1

k^3 - k^2 = k^2 - k - 12

k^3 - k^2- k^2 +k = -12

k^2(k-1) - k(k-1) = -12

(k^2 -k ) (k-1) = -12

(k)(k-1)(k-1)=-12

(k-1)(k-1)= -12/k                                                                              ...................(4)

substitute this value in equation (3)

k^2(-12/k)=36

k(-12)=36

k=36/-12

k= -3

here is your answer mark it brainly fast