Question

for what value of k,6k+1,13,5k-8 are the three consecutive terms of an AP​

Answer 1

Solution

13-(6k+1)=(5k-8)-13 {(a2-a1)=(a3-a2)=d}

13-6k-1=5k-8-13

12-6k=5k-21

12+21=5k+6k

33=11k

k=33/11

k=3

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Answer 2

Solution

Given :-

• Terms of A.P. , k , 6k + 1 ,13 ,5k - 8

Find :-

• Value of k

Explanation

We Know,

In A.P. , Common difference between any two terms will be equal.

So, We take any two terms and find Value of K.

==> (6k + 1) - k = 13 - (5k - 8)

==> 5k + 1 = 5 - 5k

==> 5k + 5k = 5 - 1

==> 10k = 4

==> k = 4/10

==> k = 2/5

==> k = 0.4

Hence

• Value of k will be = 0.4

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