The formula given below can be used to find a specific term or nth term in an arithmetic sequence.
tn = t1 + (n - 1)d
tn ----> nth term
t1 ----> 1st term
d ----> common difference
Example 1 :
The first term of an arithmetic sequence is 5 and the common difference is 7. Find the 36th term.
Solution :
Formula to find nth term of an arithmetic sequence :
tn = t1 + (n - 1)d
Substitute n = 36, t1 = 5 and d = 7.
t36 = 5 + (36 - 1)(7)
= 5 + (35)(7)
= 5 + 245
= 250
36th term is 250.
Example 2 :
The 5th term and 12th terms of an arithmetic sequence are 14 and 35 respectively. Find 56th term.
Solution :
Formula to find nth term of an arithmetic sequence :
tn = t1 + (n - 1)d
t5 = 14 t1 + (5 - 1)d = 14 t1 + 4d = 14 ----(1) |
t12 = 35 t1 + (12 - 1)d = 35 t1 + 11d = 35 ----(2) |
Solve (1) and (2) for t1 and d.
(2) - (1) :
7d = 21
Divide both sides by 7.
d = 3
Substitute d = 3 in (1).
t1 + 4(3) = 14
t1 + 12 = 14
Subtract 12 from both sides.
t1 = 2
tn = t1 + (n - 1)d
Substitute n = 56, t1 = 2 and d = 3.
t56 = 2 + (56 - 1)(3)
= 2 + (55)(3)
= 2 + 165
= 167
56th term is 167.
Example 3 :
The sum of n terms of an arithmetic sequence is 3n2 + 5n. Find the 25th term.
Solution :
Sn = 3n2 + 5n
S1 = 3(1)2 + 5(1) S1 = 3 + 5 S1 = 8 |
S2 = 3(2)2 + 5(2) S2 = 3(4) + 10 S2 = 12 + 10 S2 = 22 t1 + t2 = 22 |
S1 = 8 means sum of 1 term of the arithmetic sequence.
Sum of 1 term of an arithmetic sequence is the first term of the sequence.
t1 = 8
S2 = 22 means sum of the first two terms of the arithmetic sequence.
S2 = 22
t1 + t2 = 22
Substitute t1 = 8.
8 + t2 = 22
Subtract 8 from both sides.
t2 = 14
t1 + (2 - 1)d = 14
t1 + d = 14
Substitute t1 = 8.
8 + d = 14
Subtract 8 from both sides.
d = 6
tn = t1 + (n - 1)d
Substitute n = 25, t1 = 8 and d = 6.
t25 = 8 + (25 - 1)(6)
= 8 + (24)(6)
= 8 + 144
= 152
25th term is 152.
Example 4 :
The sum of the first two terms of an arithmetic series is 13 and the sum of the first four terms is 46. Determine the 15th term.
Solution :
Formula to find sum of first n terms of an arithmetic sequence :
Sn = (n/2)[2t1 + (n - 1)d]
S2 = 13 (2/2)[2t1 + (2 - 1)d] = 13 2t1 + d = 13 ----(1) |
S4 = 46 (4/2)[2t1 + (4 - 1)d] = 46 2[2t1 + 3d] = 46 2t1 + 3d = 23 ----(2) |
(2) - (1) :
2d = 10
Divide both sides by 2.
d = 5
Substitute d = 5 in (1).
2t1 + 5 = 13
Subtract 5 from both sides.
2t1 = 8
Divide both sides by 2.
t1 = 4
tn = t1 + (n - 1)d
Substitute n = 15, t1 = 4 and d = 5.
t15 = 4 + (15 - 1)(5)
= 4 + (14)(5)
= 4 + 70
= 74
15th term is 74.
Example 5 :
A person is employed in a company at $3000 per month and he would get an increase of $100 per year. Find the monthly salary in 24th year.
Solution :
Monthly salary in 1st year = $3000
Monthly salary in 2nd year = $3100
Monthly salary in 3rd year = $3200
3000, 3100, 3200, ..........
The above sequence is an arithmetic sequence with the first term 3000 and common difference 100.
t1 = 3000 and d = 100
tn = t1 + (n - 1)d
Substitute n = 24, t1 = 3000 and d = 100.
t24 = 3000 + (24 - 1)(100)
= 3000 + (23)(100)
= 3000 + 2300
= 5300
The monthly salary in 24th year is $5300.
Example 6 :
The mth term of an arithmetic sequence is m and nth term is m. Find the rth term.
Solution :
tn = t1 + (n - 1)d
tm = n t1 + (m - 1)d = n ----(1) |
tn = m t1 + (n - 1)d = m ----(2) |
(1) - (2) :
(m - 1)d - (n - 1)d = n - m
d[(m - 1) - (n - 1)] = n - m
d(m - 1 - n + 1) = n - m
d(m - n) = n - m
d(m - n) = -(m - n)
Divide both sides by (m - n).
d = -1
Substitute d = -1 in (1).
t1 + (m - 1)(-1) = n
t1 - m + 1 = n
t1 = m + n - 1
tn = t1 + (n - 1)d
Substitute n = r, t1 = m + n - 1 and d = -1.
tr = (m + n - 1) + (r - 1)(-1)
= m + n - 1 - r + 1
= m + n - r
rh term is (m + n - r).
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