Question

Find the general term of the A.P given by: (x + b), (x + 3b), (x + 5b),......

Answer 1

Hint: In this question it is given that we have to find the general term of the A.P given by: (x + b), (x + 3b), (x + 5b),......
so for this we need to know that if ‘a’ be the first term and ‘d’ be the common difference of an Arithmetic Progression then Then its general term is the $n^{th}$ term of the series and the $n^{th}$ term can be written as,
$$t_{n}=a+\left( n-1\right) d$$..............(1)
Complete step-by-step solution:
Given series,
 (x + b), (x + 3b), (x + 5b),......
The first term of the series a=(x + b) and the common difference,
d=(x +3b)-(x+b) =2b
Now, by equation one we can say that the general term of the given A.P ,
$$t_{n}=a+\left( n-1\right) d$$
$$=\left( x+b\right) +\left( n-1\right) \left( 2b\right) $$
$$=x+b+2bn-2b$$
$$=x+2bn-b$$
$$=x+\left( 2n-1\right) d$$
Therefore, the general term $$t_{n}=x+\left( 2n-1\right) d$$.
Note: In this question it is already given that the series follows Arithmetic progression but if they don’t mention then you have to show that the difference is common by taking consecutive three terms. i.e, if a, b, c are the first three terms then,
$$d_{1}$$=b-a, $$d_{2}$$=c-b, and if $$d_{1}$$=$$d_{2}$$ then you can say that the series is in A.P.