In the page arithmetic sequence worksheet solution2 you are going to see solution of each questions from the arithmetic sequence worksheet.
5) Find the 17th term of the A.P 4 , 9 , 14 ,…………
Solution:
First term = 4
a = 4
Common difference = 9 - 4
d = 5
n = 17
General term of an A.P
tn = a + (n - 1) d
tn = 4 + (17 - 1) (5)
= 4 + 16 (5)
= 4 + 80
= 84
Therefore 17th of A.P is 84
6) How many terms are there in the following Arithmetic progressions?
(i)-1,-5/6,-2/3,……………10/3
Solution:
First term = -1
a = -1
Common difference = t2 – t1
= (-5/6) – (-1)
= (-5/6) + 1
= (-5 + 6)/6
= 1/6
tn = 10/3
tn = a + (n - 1) d
10/3 = -1 + (n - 1) (1/6)
(10/3) + 1 = (n - 1) (1/6)
13/3 = (n - 1) (1/6)
(13/3)(6/1) = (n - 1)
26 = n -1
26 + 1= n
n = 27
Therefore 27 terms are in
the given A.P
(ii)7,13,19,……………205
Solution:
First term = 7
a = 7
Common difference = t2 – t1
= 13 – 7
= 6
tn = 205
tn = a + (n - 1) d
205 = 7 + (n - 1) (6)
205 - 7 = (n - 1) (6)
198 = (n - 1) (6)
198/6 = (n - 1)
33 = n -1
33 + 1= n
n = 34
Therefore 34 terms are in
the given A.P
(7) If 9th term of an A.P is zero, prove that its 29th term is double (twice) the 19th term
Solution:
9th term = 0
t₉ = 0
a + 8 d = 0
a = -8d
t₂₉ = a + 28 d
= -8 d + 28 d
= 20 d
= 2[10 d]
= 2 [-8d + 18 d]
= 2[a + 18 d]
t₂₉ = 2[t₁₉]
(8) The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th term
Solution:
10th term = 41
18th term = 73
a + 9 d = 41 ------- (1)
a + 17 d = 73 ------- (2)
Subtracting the second equation from first equation
a + 9 d = 41
a + 17 d = 73
(-) (-) (-)
-------------------
-8 d = -32
d = (-32)/(-8)
d = 4
Substitute d = 4 in the first equation
a + 9 (4) = 41
a + 36 = 41
a = 41 – 36
a = 5
Now we have to find 27th term
tn = a + (n - 1) d
= 5 + (27-1) 4
= 5 + 26 (4)
= 5 + 104
= 109arithmetic sequence worksheet solution2 arithmetic sequence worksheet solution2