Cinemugam (Kollywood)

Arithmetic sequence worksheet : Exercise 2.1 Questions Solution (1)The first term of an A.P is 6 and the common difference is 5. Find the A.P and its general term Solution (2) Find the common difference and 15 th term of an A.P 125 , 120 , 115 , 110 , ……….…. Solution (3) Which term of the arithmetic sequence is 24  , 23 ¼ ,22 ½ , 21 ¾ , ………. Is 3? Solution (4) Find the 12th term of the A.P √2 , 3 √2 , 5 √2 , ………… Solution (5)  Find the 17th term of the A.P 4 , 9 , 14 ,………… Solution (6) How many terms are there in the following A.P (i) -1,-5/6,-2/3,……………10/3 (ii) 7,13,19,……………205 Solution (7) If 9th term of an A.P is zero, prove that its 29th term is double (twice) the 19th term Solution (8)The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th term Solution (9) Find n so that nth terms of the following two A.P's are the same. 1,7,13,19,.............. and 100,95,90,........... Solution (10) How ma

Solution: a , b , c are in A.P Divide each term by abc, so we get (a/abc),(b/abc),(c/abc) are also an A.P By simplifying this we get, (1/bc),(1/ca),(1/ab) are also in A.P

Solution: Since a, b, c are in A.P so                       b – a = c – b                      b + b = c + a                         2 b = c + a  Taking squares on both sides                      (2 b) ² = (c + a)²                      4b² = (c + a)² Subtracting by 4ac on both sides, we get                4b² – 4ac = (c + a)² – 4 ac                4(b² – ac) = c² + a² + 2 ac – 4 ac                4(b² – ac) = c² + a² - 2 ac                4(b² – ac) = ( c – a )²

Solution: Simple interest = PNR/100                      = (25000 x 1 x 14)/100                      = 3500 Amount = principal + interest          = 25000 + 3500         = 28500 Amount at the end of the first year = 28500 Amount at the end of second year = 28500 + 3500                                                    = 32000 Amount at the end of third year = 32000 + 3500                                               = 35500 28500,32000,35500.,…………………. This is the arithmetic sequence. To find the amount of investment after 20 years we need to find 20th term tn = a +(n - 1) d a = 28500 d = 32000 – 28500                    = 3500 t20 = 28500 + (20 - 1) 3500   = 28500 + 19 (3500)   = 28500 + 66500   = 95000

Solution: m tm = n tn m [ a+ (m - 1) d ] = n [ a+ (n - 1) d ] m a + m d (m – 1)  = n a + n d (n – 1) m a + m² d – m d  = n a + n² d – n d m a - n a + m² d - n² d – m d  + n d = 0 a (m – n) + (m² - n² – m + n ) d = 0 a (m – n) + [ (m + n) (m - n) –( m - n ) ] d = 0 a (m – n) + (m – n) [ (m + n) – 1 ] d = 0 Divide by (m - n) => a + [ (m + n) – 1 ] d = 0                                  t (m + n) = 0 Therefore the (m+n) th term of the A.P is zero.

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

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

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.

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.

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