Question

If the 1st number ina deries of consecutive odd numbers is 8 less than the last number in the series, how many numbers are there in the series?

a. 7
b. 8
c. 4
d. 5
e. None​

Answer 1

Answer:

right answer is 4 hope it's help you ☺️☺️☺️☺️☺️☺️

Answer 2

Answer: There are (d) 5 numbers in this series

Step-by-step explanation:

According to the question, we have a series of odd numbers. The 1st number in the series of consecutive odd numbers is 8 less than the last number in the series.

We have to find how many odd consecutive numbers are there in this series, given the above conditions.

Step 1) Let there be x numbers. Let the last number be L.

Step 2) Let's form the series. Odd numbers are numbers which are not divisible by 2. So, in general form, any odd number can be represented by 2n+1 or 2n+3 or 2n+5 and so on...
Hence our series will be (2n+1), (2n+3), (2n+5) ... L

Step 3) Now we have to form our equation as per the condition given. Since the first number, i.e. (2n+1) is 8 less than last number, i.e. L, our equation becomes,
(2n+1)=L-8

Step 4) Solve the equation to find L.
Transposing 8 from RHS to LHS, we get,
2n+9=L

Step 5) Check the series again from step 2,
The series will go like  (2n+1), (2n+3), (2n+5) till L

Since L =  (2n+9), the final series becomes
(2n+1), (2n+3), (2n+5), (2n+7), (2n+9)

Hence we have got 5 terms in this series.

For more such questions on series, visit

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