The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals
An arithmetic progression (A.P.) is a sequence where each term differs from the previous term by a constant value. The fixed difference between consecutive terms is called common difference. The chapter explores finding nth term first term last term and sum of n terms through formulas patterns and logical reasoning. Students learn to solve real-world problems using arithmetic sequences including number patterns distance calculations financial applications and measurement-based scenarios.
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Given first three terms a₁, a₂, a₃ of A.P. are:
a₁ = 3y – 1
a₂ = 3y + 5
a₃ = 5y + 1
In an A.P., a₂ – a₁ = a₃ – a₂
So:
(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)
Simplifying LHS:
3y + 5 – 3y + 1 = 6
Simplifying RHS:
5y + 1 – 3y – 5 = 2y – 4
As LHS = RHS:
6 = 2y – 4
Adding 4 to both sides:
10 = 2y
Thus:
y = 2
Thus, 2 is the correct answer.