# Mathematics - Class 10 / Grade 10

- General Term of an Arithmetic Progression / nth Term of an Arithmetic Progression
### General Term of an Arithmetic Progression / nth Term of an Arithmetic Progression

**Tags:**Formula for the general term of an arithmetic sequence for grade 10, what is the general term? How do you find the nth term in an Arithmetic Progression Examples for class X,, nth Term of an Arithmetic Sequence Exercises for 10^{th}grade, How to find the nth term of a sequence for 10th standard, Finding the nth term of an arithmetic sequence worksheet PDF and Practice Page.General Term of an Arithmetic Progression

Formula for the nth term of the A.P. with first term

**a**and common difference**d**is**a**=_{n}**a**+ (n-1)**d**.**nth term (a**is also called the_{n})**general term of the AP**.Therefore, General Term of an A.P. = First term + (Term Number – 1) x (Common Difference)

If there are ‘

**n**’ terms in the AP, then**a**represents the_{n}**last term which is sometimes also denoted by l**.**Example1:**Find the 14^{th}, 20^{th}and nth term of the A.P. given by 9, 13, 17, 21, 25…..**Solution:**Here,**a**(first term) = 9And,

**d**= Common difference = 4 (as 13-9 = 4, 17-13=4, 21-17=4 etc.)We know that

**a**=_{n}**a**+ (n-1)**d**Therefore,

a

_{14}= a + (14-1)d= a + 13d

= 9 + 13 x 4

= 61

a

_{20}= a + (20-1)d= a + 19d

= 9 + 19 x 4

= 85

And,

a

_{n}= a+(n-1)d= 9 + (n-1) x 4

= 4n + 5

Thus, we have

a

_{14}= 61, a_{20}= 85 and a_{n}= 4n + 5

**Example 2:**The first term of an A.P. is -7 and the common difference is 5. Find its 18^{th}term and the general term.**Solution:**We have, a = first term = -7And

d = common difference = 5

Therefore, a

_{18}= a + (18-1)d= a + 17d

= -7 + 17 x 5

= 78

And,

a

_{n}= a + (n-1)x5= -7 + (n-1) x 5

= -7 + 5n -5

= 5n -12

**Example 3:**Which term of the A.P. 4, 9, 14, 19 … is 299?**Solution:**Here, first term (**a**) = 4 and common difference (**d**) = 5.Let consider 299 be the nth term of the given A.P. Then we have,

a

_{n}= 299a + (n-1)d = 299 (

**a**=_{n}**a**+ (n-1)**d**)4 + (n-1) x 5 = 299

(n-1) x 5 = 299 – 4

n-1 = 295 ÷ 5

n-1 = 59

n = 59 + 1

n= 60

Hence, 60th term of the given sequence is 299.

**Example 4:**Check whether 125 is a term of the list of numbers 3, 7, 11 ….?**Solution:**Clearly, the given sequence is an A.P. with first term**a**= 3 and common difference**d**= 4 (7-3=4, 11-7=4).Let the nth term of the given sequence be 124. Then we have,

a

_{n}= 125a + (n-1)d = 125

3 + (n-1) x 4 = 125

4n = 125

n = 125/4

Since n is not a natural number. So, 125 is not a term of the given sequence.

**Example 5:**The 5^{th}term of an A.P. is 23 and 8^{th}term is 38. Find the 12^{th }term and the general term.**Solution:**Let us consider (**a**) be the first term and (**d**) be the common difference of the given A.P.Let the A.P. be a

_{1}, a_{2}, a_{3}….. a_{n}It is given that a

_{5}= 23 and a_{8}= 38And we know that

**a**=_{n}**a**+ (n-1)**d**So, a + (5-1)d = 23

a + 4d =23 (i)

And a + (8-1)d = 38

a + 7d = 38 (ii)

Subtracting equation (ii) from equation (i), we get

-3d = -15

d = 5

Putting d =5 in equation (i), we get

a + 4d =23

a + 4 x 5 = 23

a = 23 - 20

a = 3

Therefore, a

_{12}= a + (12-1)d = 3 + 11x5 = 58 (a = 3, d = 5)And

a

_{n}= a + (n-1)d = 3 + (n-1)x5 = 5n - 2.Hence, a

_{12}= 58 and a_{n}= 5n - 2

**Example 6:**How many numbers of two digits are divisible by 5?**Solution:**We observe that 10 is the first two digit number divisible by 5 and 95 is the last two digit number divisible by 5.Thus, list of two digit numbers divisible by 5 is 10, 15, 20 ….. 95.

Clearly, it is an A.P. with first term (

**a**) = 10 and common difference (**d**) = 5 i.e. a = 10 and d = 5Let us assume there are ‘n’ terms in this A.P. Then,

nth term = 95

**a**=_{n}**a**+ (n-1)**d**95 = 10 + (n-1) x 5

95 = 10 + 5n – 5

5n = 95-5

n = 90 ÷ 5

n= 18

Hence, there are 18 numbers of two digits which are divisible by 5.

**Example 7:**Find the11th term from the last term (towards the first term) of the AP: 3, 6, 9 . . ., 111.**Solution:**Let’s write the given A.P in the reverse order such that last term becomes the first term.If we write the given AP in the reverse order, then

**a**= 111 and**d**= -3We know that

**a**=_{n}**a**+ (n-1)**d**Therefore,

a

_{11}= 111 + (11-1) x-3= 111 - 33

= 78

So, the answer is 78.

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