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
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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
2k2 – 3k + 6k – 9 = 3k2 – 7k + 6k – 14
3k2 – 7k + 6k – 14 – 2k2 + 3k – 6k + 9 = 0
k2 – 4k – 5 = 0
k2 – 5k + k – 5 = 0
k(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.