This is an important ques from the Book ML Aggarwal class 10th,, chapter – 7, ratio and proportion.

Here four numbers with a constant K are given in proportion and we have to find the value of K

Question 8, exercise 7.2

Become a Mathematics wizard and get above

Solution:It is given that

k + 3, k + 2, 3k – 7 and 2k – 3 are in proportion

We can write it as

(k + 3) (2k – 3) = (k + 2) (3k – 7)

By further calculation

2k

^{2}– 3k + 6k – 9 = 3k^{2}– 7k + 6k – 143k

^{2}– 7k + 6k – 14 – 2k^{2}+ 3k – 6k + 9 = 0k

^{2}– 4k – 5 = 0k

^{2}– 5k + k – 5 = 0k(k – 5) + 1(k – 5) = 0

(k + 1) (k – 5) = 0

So,

k + 1 = 0 or k – 5 = 0

k = -1 or k = 5

Therefore, the value of k is -1, 5.