Question

The first three terms of an A. P. are $(3y – 1), (3y + 5)$ and $(5y + 1)$. Then y equals to:

Answer 1

Hint:
We are given the three terms are given in A. P. therefore, the sum of the first and the third term will be equal to two times the second term i.e., $2 \times (3y + 5) = (3y – 1) + (5y + 1)$. Upon simplifying this equation, we will get the value of y.

Complete step by step solution:
We are given the first three terms of an A. P. as $(3y – 1), (3y + 5)$ and $(5y + 1)$.
We know that if any three terms are in an A. P. then the sum of the first and the third term is equal to twice the second term i.e., if a, b and c are in A. P. then a + c = 2b.
Therefore, we can write the given three terms using this property as:
$ \Rightarrow 2 \times (3y + 5) = (3y – 1) + (5y + 1)$
Simplifying this equation, we get
$ \Rightarrow 6y + 10 = 8y$
$ \Rightarrow 8y – 6y = 10$
$ \Rightarrow 2y = 10$
$ \Rightarrow y = \dfrac{{10}}{2} \Rightarrow 5$

Therefore, the value of y is calculated to be as 5.

Note:
Numbers are said to be in sequence when they have been arranged in a particular manner or order. Arithmetic progression is a sequence when we add a fixed number to an in number to get the next number to it. Let that fixed number is $d$ and a number in the sequence is $a$. Now to get next number to $a$, we’ll simply add $d$ in $a$ i.e. $a+d$. similarly for next $a+2d$.