HOW TO FIND THE MISSING TERM OF AN ARITHMETIC SEQUENCE

Question 1 :

Given d = 5 , S₉ = 75 find a and a₉

Solution :

S_n  =  (n/2) [2a + (n - 1)d]

By applying the values of n and d, we get

S₉  =  (9/2) [2a + (9 - 1)5]

75  =  (9/2) [2a + 40]

2a + 40  =  150/9

2a  =  -210/9

a  =  -35/3

According to the question, we find the 9th term.

9^th term :

a₉  =  a + 8d

a₉  =  (-35/3) + 8(5)

a₉  =  (-35/3) + 40

a₉  =  85/3

Hence, the values of a and a₉ are -35/3 and 85/3 respectively.

Question 2 :

Given a = 2, d = 8, S_n = 90 find n and a_n 

Solution :

S_n = 90

(n/2) [2a + (n - 1)d]  =  90

Applying the values of a and d, we get

(n/2) [2(2) + (n - 1)(8)]  =  90

(n/2) [4 + 8n - 8]  =  90

(n/2) [8n - 4]  =  90

n [4n - 2]  =  90

Dividing both sides by 2, we get

n[2n - 1]  =  45

2n² - n  - 45  =  0 

2n² - 10n + 9n - 45  =  0

2n(n - 5) + 9(n - 5)  =  0

(n - 5) (2n + 9)  =  0 

By solving for n, we will get n = 5 and  n = -9/2.

a_n  =  a + (n - 1)d

  =  2 + (n - 1)(8)

  =  2 + 8n - 8

a_n  =  8n - 6

Hence, the values of n and a_n are 5 and 8n - 6 respectively.

Question 3 :

Given a = 8, a_n = 62, S_n = 210 find n and d

Solution :

n^th term :

a_n = 62

a + (n - 1) d  =  62

By applying the value of a, we get 

8 + (n - 1) d  =  62

(n - 1)d  =  62 - 8

(n - 1)d  =  54  ----(1)

S_n = 210

(n/2) [2a + (n - 1)d]  =  210

By applying the values of a and (n - 1)d, we get

(n/2)[2(8) + 54]  =  210

(n/2)[16 + 54]  =  210

(n/2)(70)  =  210

35 n  =  210

Divide both sides by 35

n  =  210/35  =  6

By applying the value of n in (1), we get

(6 - 1) d  =  54 

5d  =  54

d  =  54/5

Hence, the values of n and d are 6 and 54/5 respectively.

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