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

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