ARITHMETIC SEQUENCE - FIRST TERM AND COMMON DIFFERENCE WORKSHEET

Problem 1 :

If 5th and 12th terms of an arithmetic sequence are 14 and 35 respectively, find the first term and common difference.

1. Answer :

Formula to find nth term of an arithmetic sequence :

tn = t1 + (n - 1)d

Given : 5th and 12th terms of an arithmetic sequence are 14 and 35 respectively.

t5 = 14

t1 + (5 - 1)d = 14

t1 + 4d = 14 ----(1)

t12 = 35

t12 + (12 - 1)d = 35

t1 + 11d = 35 ----(2)

(2) - (1) :

7d = 21

Divide both sides by 7.

d = 3

common difference = 3

Substitute d = 3 in (1).

t1 + 4(3) = 14

t1 + 12 = 14

Subtract 12 from both sides.

t1 = 2

first term = 2

Problem 2 :

Given t1 = 7 and t13 = 35. Find d and S13.

2. Answer :

t13 = 35

t1 + (13 - 1)d = 35

t1 + 12d = 35

Substitute t1 = 7.

7 + 12d = 35

Subtract 7 from both sides.

12d = 28

Divide both sides by 12.

d = 7/3

Formula to find the sum of first n terms of an arithmetic sequence :

Sn = (n/2)[2t1 + (n - 1)d]

Substitute n = 13, t1 = 7 and d = 7/3.

Problem 3 :

Given t12 = 37 and d = 3. Find t1 and S12.

3. Answer :

t12 = 37

t1 + (12 - 1)d = 37

t1 + 11d = 37

Substitute d = 3.

t1 + 11(3) = 37

t1 + 33 = 37

Subtract 33 from both sides.

t1 = 4

Formula to find the sum of first n terms of an arithmetic sequence :

Sn = (n/2)[2t1 + (n - 1)d]

Substitute n = 12, t1 = 4 and d = 3.

= (12/2)[2 ⋅ 4 + (12 - 1)(3)]

= 6[8 + (11)(3)]

= 6(8 + 33)

= 6(41)

= 246

Problem 4 :

Given t3 = 15 and S10 = 125. Find t1d and t10.

4. Answer :

t3 = 15

t1 + (3 - 1)d = 15

t1 + 2d = 15 ----(1)

S10 = 125

(10/2)[2t1 + (10 - 1)d] = 125

5[2t1 + 9d] = 125

Divide both sides by 5.

2t1 + 9d = 25 ----(2)

(2) - 2(1) :

5d = -5

Divide both sides by 5.

d = -1

Substitute d = -1 in (1).

t1 + 2(-1) = 15

t1 + -2 = 15

Add 2 to both sides.

t= 17

Formula to find nth term of an arithmetic sequence :

tn = t1 + (n - 1)d

Substitute n = 10, t1 = 17 and d = -1.

t10 = 17 + (10 - 1)(-1)

t10 = 17 + 9(-1)

t10 = 17 - 9

t10 = 8

Problem 5 :

Given d = 5 and S9 = 75. Find t1d and t9.

5. Answer :

S9 = 75

(9/2)[2t1 + (9 - 1)(5)] = 75

(9/2)[2t1 + 8(5)] = 75

(9/2)[2t1 + 40] = 75

Multiply both sides by 2/9.

2t1 + 40 = 150/9

2t1 + 40 = 50/3

Subtract 40 from both sides.

2t1 = 50/3 - 40

2t1 = (50 - 120)/3

2t1 = -70/3

Divide both sides by 2.

t1 = -35/3

Formula to find nth term of an arithmetic sequence :

tn  =  t1 + (n - 1)d

Substitute n = 9, t1 = -35/3 and d = 5.

t9 = -35/3 + (9 - 1)(5)

t9 = -35/3 + 8(5)

t9 = -35/3 + 40

t9 = (-35 + 120)/3

t9 = 85/3

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