Arithmetic Progressions Ex 5.3 Solutions
Question - 11 : - If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Answer - 11 : - 
Question - 12 : - Find the sum of the first 40 positive integers divisible by 6.
Answer - 12 : - 
Question - 13 : - Find the sum of the first 15 multiples of 8.
Answer - 13 : - 
Question - 14 : - Find the sum of the odd numbers between 0 and 50.
Answer - 14 : - 
Question - 15 : - A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:
₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc. the penalty for each succeeding day being ₹ 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?
Answer - 15 : - 
Question - 16 : - A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Answer - 16 : - 
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, eg. 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?
Answer - 17 : - 
Question - 18 : - A spiral is made upof successive semicircles, with centres alternately at A and B,
Answer - 18 : -
starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
Solution
We know,
Perimeter of a semi-circle = πr
Therefore,
P1 = π(0.5) = π/2 cm
P2 = π(1) = π cm
P3 = π(1.5) = 3π/2 cm
Where, P1, P2, P3 arethe lengths of the semi-circles.
Hence we got a series here, as,
π/2, π, 3π/2, 2π, ….
P1 = π/2 cm
P2 = π cm
Common difference, d = P2 – P1 = π– π/2 = π/2
First term = P1= a = π/2cm
By the sum of n term formula, we know,
Sn = n/2 [2a + (n –1)d]
Therefor, Sum of the length of 13 consecutivecircles is;
S13 = 13/2 [2(π/2) + (13 – 1)π/2]
= 13/2 [π + 6π]
=13/2 (7π)
= 13/2 × 7 × 22/7
= 143 cm
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 soon.
Answer - 19 : - In how many rows are the 200 logs placed and how many logs are in the top row?
Solution
We can see that the numbers of logs in rowsare in the form of an A.P.20, 19, 18…
For the given A.P.,
First term, a = 20 and common difference, d = a2−a1 =19−20 = −1
Let a total of 200 logs be placed in n rows.
Thus, Sn = 200
By the sum of nth term formula,
Sn = n/2 [2a +(n -1)d]
S12 = 12/2 [2(20)+(n -1)(-1)]
400 = n (40−n+1)
400 = n (41-n)
400 = 41n−n2
n2−41n + 400 = 0
n2−16n−25n+400 = 0
n(n −16)−25(n −16)= 0
(n −16)(n −25) = 0
Either (n −16) = 0 or n−25= 0
n = 16 or n =25
By the nth term formula,
an = a+(n−1)d
a16 = 20+(16−1)(−1)
a16 = 20−15
a16 = 5
Similarly, the 25th term couldbe written as;
a25 = 20+(25−1)(−1)
a25 = 20−24
= −4
It can be seen, the number of logs in 16th rowis 5 as the numbers cannot be negative.
Therefore, 200 logs can be placed in 16 rowsand the number of logs in the 16th row is 5.
Answer - 20 : - 
A competitor startsfrom the bucket, picks up the nearest potato, runs back with it, drops it inthe bucket, runs back to pick up the next potato, runs to the bucket to drop itin, 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?
Solution
