An important and exam oriented question from arithmetic progression chapter as it was already asked in previous year paper of 2014 in which we have been asked to find for what value of y, the terms (3y−1),(3y+5) and (5y+1) are three consecutive terms of an AP.
RS Aggarwal, class 10, chapter 5B, question no 3
Given: (3y−1),(3y+5) and (5y+1) are 3 consecutive terms of an AP.
So, difference between two consecutive terms will be same.
(3y+5)−(3y−1)=(5y+1)−(3y+5)
2(3y+5)=5y+1+3y−1
6y+10=8y
8y−6y=10
2y=10
or y=5
The value of y is 5.