## Arithmetic Sequence Worksheet Solution4

In the page arithmetic sequence worksheet solution4 you are going to see solution of each questions from the arithmetic sequence worksheet.

(12) A man has saved \$640 during the first month,\$720 in the second month and \$800 in the third month. If he continues his savings in this sequence, what will be his savings in the 25th month?

Solution:

If we write his saving like a sequence we will get 640,720,800,………… to get 25th month savings we have to find the 25th term of the sequence

a = 640   d = t2 – t1

= 720- 640

= 80

tn = a + (n-1) d

t25 = 640 + (25 - 1) 80

= 640 + 24 (80)

= 640 + 1920

= 2560

Therefore he is saving 2560 in the 25th month.

(13) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three terms.

Solution:

Let a – d , a, a + d are the first three terms.

Sum of three terms = 6

a – d + a + a + d = 6

3a = 6

a = 6/3

a = 2

Product of three terms = -120

(a – d) a (a + d) = -120

a (a²-d²) = - 120

2 (2² –d²) = -120

2(4 - d²) = -120

(4 - d²) = -120/2

(4 - d²) = -60

- d² = -60 – 4

- d² = -64

d = √64

d = ± 8

a = 2    d = 8         a = 2    d = -8

-6 , 2 , 10              10,2,-6

Therefore the three terms are -6 , 2 , 10 or 10,2,-6

(14)  The sum of three consecutive terms in an A.P whose sum is 18 and the sum of their squares is 140.

Solution:

Let a – d , a, a + d are the first three terms.

Sum of three terms = 18

a – d + a + a + d = 18

3a = 18

a = 18/3

a = 6

Sum of their squares = 140

(a – d)² + a² + (a + d)² = 140

a² + d² – 2ad + a² + a² + d² + 2ad  = 140

3 a² + 2 d² = 140

3(6)² + 2 d² = 140

3(36) + 2d² = 140

108 + 2d² = 140

2d² = 140 – 108

2d²= 32

d²= 32/2

d²= 16

d = √16

d = ± 4

d = 4       d = -4

a = 6    d = 4                        a = 6    d = -4

2,6,10                                10,6,2

Therefore the three terms are 2,6,10 or 10,6,2

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