Arithmetic Sequence Worksheet Solution3





In the page arithmetic sequence worksheet solution3 you are going to see solution of each questions from the arithmetic sequence worksheet.

(9) Find n so that the nth terms of the following two A.P’s are the same

1 , 7 ,13 ,19, ………………. and 100 , 95 , 90 ,………..

Solution:

tn = a + (n - 1) d

nth term of the first sequence

a = 1   d = t-t

              = 7-1

           d = 6

  tn = 1 + (n-1) 6

  tn = 1 + 6 n – 6

  tn = 6 n – 5   -----(1)

nth term of the second sequence

a = 100   d = t-t

                = 95 - 100

               d = -5

  tn = 100 + (n-1) (-5)

  tn = 100 - 5 n + 5

  tn = 105 - 5 n   -----(2)

(1) = (2)

6 n – 5 = 105 – 5 n

6 n + 5 n = 105 + 5

11 n = 110

      n =110/11

      n = 11

Therefore 11th terms of the given sequence are equal.


(10) How many two digit numbers are divisible by 13?

Solution:

10, 11, 12,………… 99

Now we need to find how many terms from this sequence are divisible by 13

The first two digit number divisible by 13 is 13; the next two digits number divisible by 13 is 26 and 39 so on. The last two digit numbers which are divisible by 13 is 91.

13 , 26, 39, …………….. 91

Now we need to find how many terms are there in this sequence for that let us use formula for n.

n = [(L-a)/d] + 1

a = 13    d = 26 – 13   L = 91

              d = 13

n = [(91 - 13)/13] + 1

n = (78/13) + 1

n = 6 + 1

n = 7

7 two digit numbers are divisible by 13.


(11) A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year, find the number of TVs produced in the first year and 15th year.

Solution:

Number of TVs produced in the seventh year = 1000

 Number of TVs produced in the tenth year = 1450

         t7 = 1000

         t10 = 1450

a + 6 d = 1000   ----- (1)

a + 9 d = 1450   ----- (2)

 (1) – (2)

  Subtracting second equation from first equation

                                 a + 6 d = 1000

                                 a + 9 d = 1450

                         (-)    (-)       (-)

                       ------------------

                         -3d = -450

                           d = -450/(-3)

d = 150

Substitute d = 150 in the first equation

                    a + 6(150) = 1000

                    a + 900 = 1000

                           a = 1000 -900

                           a = 100

Therefore number of TVs produced on the first year is 100

To find number of TVs produced in the 15th year year we have to find the 15th term of the A.P

tn = a + (n-1) d

t15 = 100 + (15-1) 150

t15 = 100 + 14(150)

      = 100 + 2100

      = 2200

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