Mathematics

Grade10

Easy

Question

# The 17^{th} term of an Arithmetic Sequences exceeds its 10^{th} term by 7 . Find the common difference.

- 17
- 7
- 10
- None of the above

Hint:

### The given question is about arithmetic progression. Arithmetic progression is a sequence of numbers where, the difference between two consecutive terms is constant. We are given the 17^{th} term exceeds 10^{th} term by 7. We are asked to find the common difference.

## The correct answer is: None of the above

### The first term of the progression is denoted by a_{1}.

The common difference is d.

The given statement can be written as

a_{17}– a_{10} = 7 …(1)

Common difference is the fixed difference between the consecutive numbers of the sequence. We have to add the common difference to the preceding term, to get the next term. It can be negative or positive number. It can also have value zero.

The formula for nth term of a arithmetic progression is given as follows:

a_{n} = a + (n – 1)d

Let’s see the formula for fifth and seventh term.

a_{17} = a + (17 – 1)d

= a + 16d …(2)

a_{10} = a + (10 – 1)d

= a + 9d …(3)

Using the equations (1), (2) and (3) we can write,

a_{17} – a_{10 }= (a + 16d) – (a + 9d)

7 = a + 16d – a – 9d

7 = 16d – 9d

7 = 7d

Rearranging the above equation and dividing both the sides by 7 we get,

7d = 7

d = 1

So, the common difference is 1.

For such questions, we should know the formula to find any number of the terms.