## Arithmetic Sequence Worksheet Solution2

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]

t= 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

= 109

arithmetic sequence worksheet solution2 arithmetic sequence worksheet solution2