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    • General Term of an Arithmetic Progression

      Formula for the nth term of the A.P. with first term a and common difference d is an = a + (n-1) d.

      nth term (an) is also called the 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 an represents the last term which is sometimes also denoted by l.

      Example1: Find the 14th, 20th and nth term of the A.P. given by 9, 13, 17, 21, 25…..

      Solution: Here, a (first term) = 9

      And,

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

      We know that an = a + (n-1) d

      Therefore,

      a14 = a + (14-1)d

      = a + 13d

      = 9 + 13 x 4

      = 61

      a20 = a + (20-1)d

      = a + 19d

      = 9 + 19 x 4

      = 85

      And,

      an = a+(n-1)d

      = 9 + (n-1) x 4

      = 4n + 5

      Thus, we have

      a14 = 61, a20 = 85 and an = 4n + 5



      Example 2: The first term of an A.P. is -7 and the common difference is 5. Find its 18th term and the general term.

      Solution: We have, a = first term = -7

      And

      d = common difference = 5

      Therefore, a18 = a + (18-1)d

      = a + 17d

      = -7 + 17 x 5

      = 78

      And,

      an = 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,

      an = 299

      a + (n-1)d = 299 (an = 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,

      an = 125

      a + (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 5th term of an A.P. is 23 and 8th term is 38. Find the 12th 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 a1, a2, a3….. an

      It is given that a5 = 23 and a8 = 38

      And we know that an = 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, a12 = a + (12-1)d = 3 + 11x5 = 58 (a = 3, d = 5)

      And

      an = a + (n-1)d = 3 + (n-1)x5 = 5n - 2.

      Hence, a12 = 58 and an = 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 = 5

      Let us assume there are ‘n’ terms in this A.P. Then,

      nth term = 95

      an = 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 = -3

      We know that an = a + (n-1) d

      Therefore,

      a11 = 111 + (11-1) x-3

      = 111 - 33

      = 78

      So, the answer is 78.

       

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