HOW TO FIND THE TERMS POSITION IN ARITHMETIC SEQUENCE

The given number is which term of the arithmetic sequence

Problem 1 :

Which term of the AP : 3, 8, 13, 18, …is 78 ?

Solution :

The given AP is 3, 8, 13, 18, … is 78

Here, a = 3, d = 8 – 3 ==> d = 5

t_n = 78

t_n = a + (n – 1)d

78 = 3 + (n – 1)5

78 = 3 + 5n – 5

78 = -2 + 5n

78 + 2 = 5n

80 = 5n

80/5 = n

16 = n

16^th term of the AP is 78.

Problem 2 :

Which term of the AP : 3,15,27, 39, … will be 120 more than its 21^st term?

Solution :

The given AP is 3, 15, 27, 39, … is 120

Here, a = 3, d = 15 – 3

d = 12

120 more than its 21^st term

a_n = 120 + a₂₁

a_n = 120 + a + 20d

a_n = 120 + 3 + 20(12)

a_n = 120 + 3 + 240

a_n = 363

a_n = a + (n – 1)d

363 = 3 + (n – 1)12

363 = 3 + 12n – 12

372 = 12n

n = 31

Therefore, the 31^st term of the given AP is more than the 21^st term. 

Problem 3 :

The 17^th term of an A.P. exceeds its 10^th term by 7, Find the common difference.

Solution :

a₁₇ = a₁₀ + 7

a + 16d = (a + 9d) + 7

16d - 9d = 7

7d = 7

d = 1

So, the common difference is 1.

Problem 4 :

Find whether - 150 is a term of the AP 17, 12, 7, 2, … ?

Solution :

Here, a = 17, d = 12 – 17 ==> d = -5

t_n = -150

17 + (n – 1) × -5  = -150

17 – 5n + 5 = -150

22 – 5n = -150

-150 – 22 = -5n

-172 = -5n

172/5 = n

34.4 = n

So, -150 is not a term of the given AP.

Problem 5 :

170 is the n^th term of -4, 2, 8, .............

Solution :

t_n = 170

a + (n - 1)d = 170

a = -4, d = 2 + 4 ==> 6

-4 + (n - 1) 6 = 170

(n - 1)6 = 170 + 4

(n - 1) 6 = 174

n - 1 = 174 / 6

n - 1 = 29

n = 30

So, 30^th term of the sequence is 170.

Problem 6 :

97 is the n^th term of -3, 1, 5, ......

Solution :

t_n = 97

a = -3, d = 1 + 3 ==> 4

-3 + (n - 1) 4 = 97

4(n - 1) = 100

n - 1 = 25

n = 25 + 1

n = 26