0 deepaksoniGuru Asked: March 18, 20212021-03-18T11:27:10+05:30 2021-03-18T11:27:10+05:30In: ICSE If k + 3, k + 2, 3k – 7 and 2k – 3 are in proportion, find k. 0 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 ml aggarwal solutionratio and proportion Share Facebook 1 Answer Voted Oldest Recent AAREKH Guru 2021-03-20T11:55:10+05:30Added an answer on March 20, 2021 at 11:55 am 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 – 14 3k^{2} – 7k + 6k – 14 – 2k^{2} + 3k – 6k + 9 = 0 k^{2} – 4k – 5 = 0 k^{2} – 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. 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions Question 16. In the given figure, chords AB and CD of the circle are produced to meet at ... Question 15. (a) Prove that a cyclic parallelogram is a rectangle. (b) Prove that a cyclic rhombus is ... Question 14. In the given figure, ABC is an isosceles triangle in which AB = AC and circle ... Ques 13(b) In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a ... Question 13. (a) In the figure (i) given below, ED and BC are two parallel chords of the ...

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.