Arithmetic Progressions Ex 5.2 Solutions
Question - 11 : - Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
Answer - 11 : - 
Question - 12 : - Two APs have thesame common difference. The difference between their 100th termis 100, what is the difference between their 1000th terms?
Answer - 12 : - 
Question - 13 : - How many three digit numbers are divisible by 7?
Answer - 13 : -
First three-digit number that is divisible by7 are;
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
Therefore, 105, 112, 119, …
All are three digit numbers are divisible by 7and thus, all these are terms of an A.P. having first term as 105 and commondifference as 7.
As we know, the largest possible three-digitnumber is 999.
When we divide 999 by 7, the remainder will be5.
Therefore, 999-5 = 994 is the maximum possiblethree-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
first term, a = 105
common difference, d = 7
an = 994
n = ?
As we know,
an = a+(n−1)d
994 = 105+(n−1)7
889 = (n−1)7
(n−1) = 127
n = 128
Therefore, 128 three-digitnumbers are divisible by 7.
Question - 14 : - How many multiples of 4 lie between 10 and 250?
Answer - 14 : -
The first multiple of 4 that is greater than10 is 12.
Next multiple will be 16.
Therefore, the series formed as;
12, 16, 20, 24, …
All these are divisible by 4 and thus, allthese are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows, now;
12, 16, 20, 24, …, 248
Let 248 be the nth termof this A.P.
first term, a = 12
common difference, d = 4
an = 248
As we know,
an = a+(n−1)d
248 = 12+(n-1)×4
236/4 = n-1
59 = n-1
n = 60
Therefore, there are 60multiples of 4 between 10 and 250.
Question - 15 : - For what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?
Answer - 15 : - 
Question - 16 : - Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
Answer - 16 : -
Given,
Third term, a3 = 16
As we know,
a +(3−1)d =16
a+2d = 16………………………………………. (i)
It is given that, 7th termexceeds the 5th term by 12.
a7 − a5 = 12
[a+(7−1)d]−[a +(5−1)d]= 12
(a+6d)−(a+4d) = 12
2d = 12
d = 6
From equation (i), we get,
a+2(6) = 16
a+12 = 16
a = 4
Therefore, A.P. will be4, 10, 16, 22, …
Question - 17 : - Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Answer - 17 : - 
Question - 18 : - The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Answer - 18 : - 
Question - 19 : - Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000 ?
Answer - 19 : - 
Question - 20 : - Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n_th week, her weekly savings become Rs 20.75, find n.
Answer - 20 : -
Given that, Ramkali saved Rs.5 in first weekand then started saving each week by Rs.1.75.
Hence,
First term, a = 5
and common difference, d = 1.75
Also given,
an = 20.75
Find, n = ?
As we know, by the nth termformula,
an = a+(n−1)d
Therefore,
20.75 = 5+(n -1)×1.75
15.75 = (n -1)×1.75
(n -1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n -1 = 9
n = 10
Hence, n is 10.