# Arithmetic Sequences & Series 2 ```Te Hiringa i te Mahara
Arithmetic Sequences and Series 2
Kia mahara ake:
The nth term in an arithmetic sequence is illustrated by t n  a  (n  1)d where a is the 1st term
in the sequence and d is the fixed difference (common difference) between consecutive
terms.
The sum for an arithmetic sequence to the nth term is illustrated by S n  n2 [2a  (n  1)d ] ,
where a is the 1st term in the sequence and d is the fixed difference (common difference)
between consecutive terms; and n is the number of terms.
Another equivalent formula is S n  n2 [a  l ] , where l is the last term.
Hei Mahi (I)
(a)
Find the sum for the first 10 terms in the series
1) Show that the sequence tn:{0.9, 0.7, 0.5, …} is an arithmetic
sequence.
2) Determine the first 3 terms of the sequence
tn = 2n6, n  {1, 2, 3, …}.
3) Find the rule for the arithmetic sequence
1 3 5
tn:{ , , , …}.
8 8 8
4) Find the 10th term of the arithmetic sequence tn:{10, 6, 2, …}.
5) The 3 consecutive terms of an arithmetic sequence are 3.6, y, 8.2. Find
the value of y.
6) Insert 3 evenly spaced numbers between -2 and 10.
7) Find the 10th term of the arithmetic sequence where the first term is 5
and the 4th term is 17.
8) Find the difference between the 4th term and the 10th term of the
arithmetic sequence
tn:{2, 1, 4, …}.
9) The 10th term in an arithmetic sequence is 8 and the 4th term is -4.
Determine the first term a.
10) Find the sum to first 8 terms of the arithmetic sequence tn:{14, 12, 10,
…}.
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NGĀ WHAKAUTU
Hei Mahi (I)
(a)
1)
t 2  t1  0.7  0.9  0.2
t 3  t 2  0.5  0.7  0.2
Common difference = 0.2
Thus, it is an arithmetic sequence.
2) t1  2  1  6  4
t 2  2  2  6  2
t3  2  3  6  0
The first 3 terms are 4, 2, and 0.
3)
1
2 1
a ,d 
8
8 4
1
1
t n   (n  1) 
8
4
1 n 1
tn   
8 4 4
n 1
tn  
4 8
This is the required rule.
4)
a  10, d  4
t10  10  (10  1)  4
t10  10  9  4
t10  26
The 10th term is 26.
5)
(3.6  8.2)
y
2
y  5.9
6)
a  2, t 5  10
 2  (5  1)d  10
 2  4d  10
4d  12
12
d 
4
d 3
Thus we have -2, 1, 4, 7, 10 evenly spaced.
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7)
8)
 5, t 4  17
 17
 17
 12
12
d 
3
d 4
Now find t10 :
t10  5  (10  1)  4
t10  5  9  4
t10  41
a  2, d  3
t 4  2  (4  1)   3
a
5  (4  1)d
5  3d
3d
t4  2  3   3
t 4  7
t10  2  (10  1)  3
t10  2  9  3
t10  25
9)
Difference  7   25
 18
a  (10  1)d  t10
a  9d  8 
a  (4  1)d  4
a  3d  4 
Subtract  from 
6d  12
12
6
d 2
Now replace d in 
a  9 2  8
d
a  18  8
a  8  18
a  10
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10)
a  14, d  2
8
S 8  [2   14  (8  1)  2]
2
S 8  4[28  14]
S 8  4   14
S 8  56
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