Problem 1 :
If 5^{th} and 12^{th} terms of an arithmetic sequence are 14 and 35 respectively, find the first term and common difference.
1. Answer :
Formula to find n^{th} term of an arithmetic sequence :
t_{n} = t_{1} + (n - 1)d
Given : 5^{th} and 12^{th} terms of an arithmetic sequence are 14 and 35 respectively.
t_{5} = 14 t_{1} + (5 - 1)d = 14 t_{1} + 4d = 14 ----(1) |
t_{12} = 35 t_{12} + (12 - 1)d = 35 t_{1} + 11d = 35 ----(2) |
(2) - (1) :
7d = 21
Divide both sides by 7.
d = 3
common difference = 3
Substitute d = 3 in (1).
t_{1} + 4(3) = 14
t_{1} + 12 = 14
Subtract 12 from both sides.
t_{1} = 2
first term = 2
Problem 2 :
Given t_{1} = 7 and t_{13} = 35. Find d and S_{13}.
2. Answer :
t_{13} = 35
t_{1} + (13 - 1)d = 35
t_{1} + 12d = 35
Substitute t_{1} = 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 :
S_{n} = (n/2)[2t_{1} + (n - 1)d]
Substitute n = 13, t_{1} = 7 and d = 7/3.
Problem 3 :
Given t_{12} = 37 and d = 3. Find t_{1} and S_{12}.
3. Answer :
t_{12} = 37
t_{1} + (12 - 1)d = 37
t_{1} + 11d = 37
Substitute d = 3.
t_{1} + 11(3) = 37
t_{1} + 33 = 37
Subtract 33 from both sides.
t_{1} = 4
Formula to find the sum of first n terms of an arithmetic sequence :
S_{n} = (n/2)[2t_{1} + (n - 1)d]
Substitute n = 12, t_{1} = 4 and d = 3.
= (12/2)[2 ⋅ 4 + (12 - 1)(3)]
= 6[8 + (11)(3)]
= 6(8 + 33)
= 6(41)
= 246
Problem 4 :
Given t_{3} = 15 and S_{10} = 125. Find t_{1}, d and t_{10}.
4. Answer :
t_{3} = 15
t_{1} + (3 - 1)d = 15
t_{1} + 2d = 15 ----(1)
S_{10} = 125
(10/2)[2t_{1} + (10 - 1)d] = 125
5[2t_{1} + 9d] = 125
Divide both sides by 5.
2t_{1} + 9d = 25 ----(2)
(2) - 2(1) :
5d = -5
Divide both sides by 5.
d = -1
Substitute d = -1 in (1).
t_{1} + 2(-1) = 15
t_{1} + -2 = 15
Add 2 to both sides.
t_{1 }= 17
Formula to find n^{th} term of an arithmetic sequence :
t_{n} = t_{1} + (n - 1)d
Substitute n = 10, t_{1} = 17 and d = -1.
t_{10} = 17 + (10 - 1)(-1)
t_{10} = 17 + 9(-1)
t_{10} = 17 - 9
t_{10} = 8
Problem 5 :
Given d = 5 and S_{9} = 75. Find t_{1}, d and t_{9}.
5. Answer :
S_{9} = 75
(9/2)[2t_{1} + (9 - 1)(5)] = 75
(9/2)[2t_{1} + 8(5)] = 75
(9/2)[2t_{1} + 40] = 75
Multiply both sides by 2/9.
2t_{1} + 40 = 150/9
2t_{1} + 40 = 50/3
Subtract 40 from both sides.
2t_{1} = 50/3 - 40
2t_{1} = (50 - 120)/3
2t_{1} = -70/3
Divide both sides by 2.
t_{1} = -35/3
Formula to find n^{th} term of an arithmetic sequence :
t_{n} = t_{1} + (n - 1)d
Substitute n = 9, t_{1} = -35/3 and d = 5.
t_{9} = -35/3 + (9 - 1)(5)
t_{9} = -35/3 + 8(5)
t_{9} = -35/3 + 40
t_{9} = (-35 + 120)/3
t_{9} = 85/3
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