Question

For what value of k,2k+7,k+5 and 3k+2 consecutive terms of an ap.hence find the terms

Answer 1

Answer:

k=1/3

and the terms are 7.6 , 5.3 , 3 with d=2.3

Step-by-step explanation:

as they are consecutive terms

let 1st term be 2k+7

2nd term be k+5

3rd term be 3k+2

then a = 2k+7 ( as 1st term)

then 2nd term can be written as

k+5 = a+d where a= 2k+7( as 1st term)

then, k+5 = 2k+7+ d

-k-2=d..... (1)

3rd term can also be written as

3k+2= a+2d

where a =2k+7

3k+2= 2k+7 +2d

(k-5) /2=d......(2)

equating (1) and (2)

-k-2=(k-5) /2

k=1/3

putting the value of k in term 1st ,2nd and 3rd

we get 1st term 2/3+7=7.6

2nd term = 1/3+5 = 5.3

3rd term = 3/3+2 = 3

we can see the d = 2.3