HOW TO FIND THE TOTAL NUMBER OF TERMS IN AN ARITHMETIC SEQUENCE

Problem 1 :

Find the number of terms in the following arithmetic sequence.

7, 13, 19, ............. , 205

Solution :

t₁ = 7

d = t₂ - t₁ 

= 13 - 7

= 6

l = 205

Formula to find the number of terms in an arithmetic sequence :

Substitute l = 205, t₁ = 7 and d = 6.

Problem 2 :

Find the number of terms in the following arithmetic sequence.

-1,-5/6,-2/3, ............., 10/3

Solution :

t₁ = -1

d = t₂ - t₁ 

= -5/6 - (-1)

= -5/6 + 1

= (-5 + 6)/6

= 1/6

l = 10/3

Formula to find the number of terms in an arithmetic sequence :

Substitute l = 10/3, t₁ = -1 and d = 1/6.

Some Other Problems on Arithmetic Sequence

Problem 3 :

The 10^th and 18^th terms of an arithmetic sequence are 41 and 73 respectively. Find the 27^th term.

Solution :

Formula to find nth term of an arithmetic sequence :

t_n = t₁ + (n - 1)d

(2) - (1) :

8d = 32

Divide both sides by 8.

d = 4

Substitute d = 4 in (1).

t₁ + 9(4) = 41

t₁ + 36 = 41

Subtract 36 from both sides.

t₁ = 5

t_n = t₁ + (n - 1)d

Substitute n = 27, t₁ = 5 and d = 4.

t₂₇ = 5 + (27 - 1)(4)

= 5 + (26)(4)

= 5 + 104

= 109

27^th term is 109.

Problem 4 :

Find n so that the n^th terms of the following two arithmetic sequences are the same.

1, 7, 13, 19, ............. 

and

100, 95, 90, .............

Solution :

1, 7, 13, 19, .............

In the arithmetic sequence above,

t₁ = 1

d = t₂ - t₁

= 7 - 1

= 6

t_n = t₁ + (n - 1)d

Substitute t₁ = 1 and d = 6.

t_n = 1 + (n - 1)(6)

t_n = 1 + 6n - 6

t_n = 6n - 5 ----(1)

100, 95, 90, .............

In the arithmetic sequence above,

t₁ = 100

d = t₂ - t₁

= 95 - 100

= -5

t_n = t₁ + (n - 1)d

Substitute t₁ = 100 and d = -5.

t_n = 100 + (n - 1)(-5)

t_n = 100 - 5n + 5

t_n = 105 - 5n ----(2)

From the given information,

(1) = (2)

t_n = t_n

6n - 5 = 105 - 5n

Add 5n to both sides.

11n - 5 = 105

Add 5 to both sides.

11n = 110

Divide both sides by 11.

n = 10

Problem 5 :

Using arithmetic sequence, find the number of two-digit numbers which are evenly divisible by 13.

Solution :

Two-digit numbers are

10, 11, 12,………… 99

The first two-digit number which is evenly divisible by 13 is 13.

After 13, to find the next two-digit number divisible by 7, we have to add 13 to 13. So, the second two-digit number which is evenly divisible by 13 is 26.

In this way, to get the succeeding two-digit numbers which are evenly divisible by 13, we just have to add 13 as given below. 

13, 26, 39,…………

The last two-digit number which is evenly divisible by 13 is 91.

13, 26, 39,…………, 91

Clearly, the above sequence of two-digit numbers divisible by 13 forms an arithmetic sequence with common difference 13.

Formula to find the number of terms in an arithmetic sequence :

Substitute l = 91, t₁ = 13 and d = 13.

There are 7 two-digit numbers which are evenly divisible by 13.

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