Arithmetic Progression
Mathematics Class Tenth
NCERT Exercise 5.3 Q12-20
Question. (12) Find the sum of the first 40 positive integers divisible by 6.
Solution:
Here, the first positive integers divisible by 6 = 6
Thus, here, First term, `a = 6`
Common difference, `d = 6`
Number of terms `n = 40`
Therefore, `S_40=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`S_40=40/2[2xx6+(40-1)6]`
`=20[12+39xx6]`
`=20[12+234]`
`=20xx246`
`=>S_40=4920`
Thus, sum of first 40 integers divisible by 6 = 4920 Answer
Question. (13) Find the sum of the first 15 multiples of 8.
Solution:
Here, first multiple of 8, i.e. First term, `a = 8`
Common difference, `d = 8`
Total number of terms `n = 15`
Therefore, `S_15 = ?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`S_15 = 15/2[2xx8+(15-1)8]`
`=15/2[16+14xx8]`
`=15/2[16+112]`
`=15/2xx128`
`=15xx64`
`S_15 = 960`
Thus, sum of first 15 multiples of 8 = 960. Answer
Question. (14) Find the sum of the odd numbers between 0 and 50`
Solution:
Here, first odd number between 0 and 50 = 1
This means, First term, `a= 1`
Common difference, `d = 2`
Total number of terms, `n = 50/2 = 25`
Therefore, Sum of first 25 terms `S_25=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_25 = 25/2[1xx2+(25-1)2]`
`=25/2[2+24xx2]`
`=25/2[2+48]`
`=25/2xx50`
`=25xx25`
`S_25 = 625`
Thus, sum of the odd numbers between 0 and 50 = 625. Answer
Question. (15) A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
Here, given, First term, `a = 200`
Common difference, `d = 50`
Total number of terms, `n = 30`
Therefore, sum of first 30 terms, `S_30=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_30=30/2[2xx200+(30-1)50]`
`=15[400+29xx50]`
`=15[400+1450]`
`=15xx1850`
`=>S_30=27750`
Thus, contractor has to pay Rs 27750 as penalty. Answer
Question. (16) A sum of Rs 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 20 less that its preceding prize, find the value of each of the prizes.
Solution:
Here, total number of terms, `n=7`
Common difference, `d = 20`
And Sum of first `7` terms `S_7=700`
Thus, First term `a` and other six terms =?
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_7 = 7/2[2a+(7-1)20]`
`=>700=7/2[2a+6xx20]`
`=>700=7/2[2a+120]`
`=>700=7/2xx2[a+60]`
`=>700=7[a+60]`
`=>a+60=700/7`
`=>a+60=100`
`:. a = 100-60=40`
Now, we know that `n^(th)` term `a_n=a+(n-1)d`
`:. a_2 = 40+(2-1)20`
`=>a_2=40+20=60`
Similarly, `a_3 = 40+(3-1)20`
`=>a_3=40+2xx20=40+40`
`=>a_3 = 80`
Similarly, `a_4 = 40+(4-1)20`
`=>a_4 = 40+3xx20=40+60`
`=>a_4 = 100`
Similarly, `a_5 = 40+(5-1)20`
`=>a_5 = 40+4xx20=40+80`
`=>a_5 = 120`
Similarly, `a_6 = 40+(6-1)20`
`=>a_6 = 40+5xx20=40+100`
`=>a_6=140`
Similarly, `a_7=40+(7-1)20`
`=>a_7 = 40+6xx20=40+120`
`=>a_7 = 160`
Thus, values of each of the prizes are, 160, 140, 120, 100, 80, 60, and 40 respectively. Answer
Question. (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 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Here, given, there are 3 sections in each of the class
Therefor, number of trees planted by Class I `= 1xx3=3`
Similarly, no. of trees planted by Class II `= 2xx3=6`
Similarly, no. of trees planted by Class III `= 3xx3=9`
And total number of classes = 12
Thus, here we have, First term `a = 3`
Common difference, `d = 3`
Total number of terms `n = 12`
Therefore, sum of first `12` terms, i.e. `S_12=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_12 = 12/2[2xx3+(12-1)3]`
`=6[6+11xx3]`
`=6[6+33]`
`=6xx39`
`=>S_12= 234`
Thus, total number of trees planted by students will be = 234. Answer
Question. (18) A spiral is made up of succeessve semicircles, with centres alternatedly at A and B, starting with centre at A, of radii `0.5cm, 1.0cm,` `1.5cm, 2.0cm`, ... as shown in Figure. What is the total lenght of such spiral made uup of thirteen consecutive semicircles?(Take `pi=22/7`)
[Hint: Lenght of succeesive semicircles is `l_1, l_2, l_3, l_4,`.... with centres at A, B, A, B,.... respectively.]
Solution:
Here, given, radii semicircles are `0.5cm, 0.1, 1.5cm, 2.0 cm,` .... respectively.
We know that, Circumference of a semicircle `=pi\ r`
Therefor, circumference, i.e. lenght of 1st semicircle, `l_1 = 0.5pi`
Similarly, circumference of 2nd semicircle ` `l_2 = 1pi=pi`
Similarly, circumference of 3rd semicircle, `l_3=1.5pi` And so on
Given, total number of circles = 13
Thus, here we have, First term, `a=0.5pi`
Common difference, `d = 0.5`
Total no. of semicircles, i.e. total number of terms, `n = 13`
Sum of first `n=13` terms, i.e. `S_13=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_13 = 13/2[2xx0.5pi+(13-1)0.5pi]`
`=13/2\pi[2xx0.5+12xx0.5]`
`=13/2\2xxpi[0.5+6xx0.5]`
`=13pi[0.5+3]`
`=13pixx3.5`
`=13xx3.5xx22/7`
`S_13=143`
Thus, total lenght of all spirals `= 143\ cm` Answer
Question. (19) 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (See figure). In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:
Here given, First term, `a = 20`
Common difference, `d = -1`
Sum of first `n^(th)` terms `S_n = 200`
Therefore, `n=?`
And, `a_n=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`:. S_n = n/2[2xx20+(n-1)-1]`
`=>200 = n/2[40-n+1]`
`=>200xx2=n[41-n]`
`=>400= -n^2+41n`
`=>-n^2+41n-400=0`
`=>n^2-41n+400=0`
`=>n^2-16n-25n+400=0`
`=>n(n-16)-25(n-16)`
`=>(n-25)(n-16)=0`
Now, if `n-25=0`
`:. n = 25`
And if `n-16=0`
`:. n = 16`
Thus, `n=25 or n =16`
Now, we know that `n^(th)` term `a_n=a+(n-1)d`
If `n=25`
`:. a_25 = 20+(25-1)-1`
`=20+24xx(-1)=20-24=-4`
Since, it is negative value, which is not possible in the case of placement of logs, thus, it is discarded
Now, if `n=16`
`:. a_16=20+(16-1)-1`
`=20+15xx(-1)=20-15`
`a_16=5`
Thus, total number of rows = 16 and number of logs in the top row = 5. Answer
Question. (20) In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten (10) potatoes in the line (see fig).
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops 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 meters) run by competitor is `2xx5+2xx(5+3)`]
Solution:
Given, Distance of 1st potato from the bucket = 5m
∴ Distance covered to drop 1st potato `= 5m xx 2 =10m`
Distance of 2nd potato from the bucket `= 5+3=8m`
∴ Distance covered to drop 2nd potato `= 8m xx2 = 16m`
Distance of 3rd potato from the bucket `= 8+3 = 11m`
∴ Distance covered to drop the 3 rd potato `=11m xx2 = 22m`
And so on ......
List of numbers formed from above process is `10m, 16m, 22m`, .....
Thus, here, First term, `a= 10`
Common difference, `d = 6`
Total number of terms, `n=10`
∴ Sum of first 10 terms `S_10=?`
We know that, Sum of first `n` terms `S_n=n/2[2a+(n-1)d]`
`S_10=10/2[2xx10+(10-1)6]`
`=5[20+9xx6]`
`=5[20+54]=5xx74`
`=>S_16=370`
Thus, total distance run by competitor = 370m. Answer
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