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Q : 3      In the following APs, find the missing terms in the boxes :

             (v)    \small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22

Given AP series is Here,  Now, Now, we know that Now, And  And  And  Therefore, missing terms are   53,23,8,-7  AP series is 53,38,23,8,-7,-22

Q : 3     In the following APs, find the missing terms in the boxes :

             (iv)   \small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6

Given AP series is Here,  Now, we know that Now, And  And  And  Therefore, missing terms are   -2,0,2,4  AP series is -4,-2,0,2,4,6

Q : 3     In the following APs, find the missing terms in the boxes :

             (iii)   \small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}

Given AP series is Here,  Now, we know that Now, And  Therefore, missing terms are   and 8  AP series is 

Q : 3     In the following APs, find the missing terms in the boxes :

            (ii)   \small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3

Given AP series is Here,  Now, Now, we know that Now, And  Therefore, missing terms are  18 and 8  AP series is 18,13,8,3

Q : 3     In the following APs, find the missing terms in the boxes :

           (i)  \small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26

Given AP series is Here,  Now, we know that Now, Therefore, the missing term is 14

Q : 20   In a potato race, a bucket is placed at the starting point, which is \small 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. \small 5.6).

              

 

 

            A competitor starts from the bucket, picks up the nearest potato, runs back with it,  drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What  is the total distance the competitor has to run?
            [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is  \small 2\times5+2\times (5+3)  ]

Distance travelled by the competitor in picking and dropping 1st potato  Distance travelled by the competitor in picking and dropping 2nd potato  Distance travelled by the competitor in picking and dropping 3rd potato  and so on we can clearly see that it is an AP with first term (a) = 10  and common difference(d) = 6 There are 10 potatoes in the line  Therefore, total distance travelled by...

Q : 19      \small 200  logs are stacked in the following manner: \small 20 logs in the bottom row, \small 19  in the next row, \small 18 in the row next to it and so on (see Fig. \small 5.5). In how many rows are the \small 200 logs placed and how many logs are in the top row?

               

As, the rows are going up, the no of logs are decreasing,  We can clearly see that 20, 19, 18, ..., is an AP . and here    Let suppose 200 logs are arranged in 'n' rows, Then,  Now, case (i) n = 25 But number of rows can not be in negative numbers  Therefore, we will reject the value n = 25 case (ii) n = 16 Therefore, the number of rows in which 200 logs are arranged is equal...

Q : 18       A spiral is made up of successive semicircles, with centres alternately at \small A and \small B​​​​​​, starting with centre at \small A, of radii \small 0.5\hspace {1mm}cm,1.0\hspace {1mm}cm,1.5\hspace {1mm}cm,2.0\hspace {1mm}cm,...  as shown in Fig. \small 5.4. What is the total length of such a spiral made up of thirteen  consecutive semicircles? (Take   \pi =\frac{22}{7}  )

               

                [Hint : Length of successive semicircles is  \small l_1,l_2,l_3,l_4,...  with centres at  \small A,B,A,B,..., respectively.]

From the above given figure Circumference of 1st semicircle Similarly, Circumference of 2nd semicircle  Circumference of 3rd semicircle  It is clear that this is an AP with  Now, sum of length of 13 such semicircles is given by Therefore, sum of length of 13 such semicircles is 143 cm 

Q : 17       In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I​​​​ will plant \small 1 tree, a section of Class II will plant \small 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

First there are 12 classes and each class has 3 sections Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1.Thus every class will plant 3 times the number of their class  Similarly, No. of trees planted by 3 sections of class 1 = 3 No. of trees planted by 3 sections of class 2 = 6 No. of trees planted by 3 sections of class 3 = 9 No. of trees...

Q : 16         A sum of Rs \small 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs \small 20 less than its preceding prize, find the value of each of the prizes.

It is given that Each price is decreased by 20 rupees,  Therefore,  d = -20 and there are total 7 prizes so n = 7 and sum of prize money is Rs 700 so  Let a be the prize money given to the 1st student Then, Therefore, the prize given to the first student is Rs 160 Now, Let  is the prize money given to the next 6 students then,    Therefore, prize money given to 1 to 7 student...

Q : 15      A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs \small 200  for the first day, Rs \small 250 for the second day, Rs \small 300 for the third day, etc., the penalty for each succeeding day being Rs \small 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by \small 30 days?

It is given that Penalty for delay of completion beyond a certain date is Rs   for the first day, Rs  for the second day, Rs  for the third day and  penalty for each succeeding day being Rs  more than for the preceding day We can clearly see that 200,250,300,..... is an AP  with  Now, the penalty for 30 days is given by the expression Therefore, the penalty for 30 days is 27750

Q :  14        Find the sum of the odd numbers between \small 0 and \small 50.

Odd number between 0 and 50 are   1,3,5,...49 This is an AP with  There are total 25 odd number between 0 and 50 Now, we know that  Therefore,  sum of the odd numbers between and   625

Q : 13     Find the sum of the first \small 15  multiples of  \small 8.

First 15 multiples of 8 are   8,16,24,... This is an AP with  Now, we know that  Therefore,  sum of the first 15 multiple of 8 is 960

Q : 12    Find the sum of the first \small 40 positive integers divisible by  \small 6.

Positive integers divisible by 6 are  6,12,18,... This is an AP with  Now, we know that  Therefore,  sum of the first positive integers divisible by   is 4920

Q : 11      If the sum of the first \small n terms of an AP is  \small 4n-n^2, what is the first term (that is  \small S_1)? What is the sum of first two terms? What is the second term? Similarly, find the \small 3rd, the \small 10th and the \small nth terms

It is given that  the sum of the first terms of an AP is   Now, Now, first term is  Therefore, first term is 3 Similarly, Therefore, sum of first two terms is 4 Now, we know that Now, Similarly,  

Q : 10    Show that  \small a_1,a _2,...,a_n,... form an AP where an is defined as below :

               (ii) \small a_n=9-5n 

               Also find the sum of the first \small 15 terms in each case.

It is given that  We will check values of  for different values of n and so on. From the above, we can clearly see that this is an AP with the first term(a) equals to 4 and common difference (d) equals to -5 Now, we know that  Therefore, the sum of 15 terms is -465

Q : 10   Show that  \small a_1,a_2,...,a_n,...  form an AP where an is defined as below :

               (i)   \small a_n=3+4n

               Also find the sum of the first \small 15 terms .

It is given that  We will check values of  for different values of n and so on. From the above, we can clearly see that this is an AP with the first term(a) equals to 7 and common difference (d) equals to 4 Now, we know that  Therefore, the sum of 15 terms  is  525

Q : 9    If the sum of first \small 7 terms of an AP is \small 49 and that of \small 17 terms is \small 289 , find the sum of
            first \small n terms.

It is given that Now, we know that  Similarly, On solving equation (i) and (ii) we will get  a = 1 and d = 2 Now, the sum of first n terms is  Therefore, the sum of n terms  is  

Q : 8      Find the sum of first \small 51 terms of an AP whose second and third terms are \small 14 and \small 18
              respectively.

It is given that And  Now, Now, we know that  Therefore, there are 51 terms  and their sum is 5610

Q : 7     Find the sum of first \small 22 terms of an AP in which \small d=7 and \small 22nd term is \small 149.

It is given that Now, we know that Now, we know that  Therefore, there are 22 terms  and their sum is 1661
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