  # Important Questions for CBSE Class 10 Maths Chapter 5 - Arithmetic Progressions 3 Mark Question

## CBSE Class 10 Maths Chapter-5 Arithmetic Progressions – Free PDF Download

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CBSE Class 10 Maths Chapter-5 Arithmetic Progressions Important Questions

## 3 Mark Questions

1. Write first four terms of the AP, when the first term a and common difference d are given as follows:
(i) a = 10, d=10
(ii) a= -2, d=0
(iii) a=4, d=-3
(iv) a=-1, d= ½
(v) a=-1.25, d=-0.25
Ans. (i) First term = a =10, d=10
Second term = a+d = 10 + 10 = 20
Third term = second term + d = 20 + 10 = 30
Fourth term = third term + d = 30 + 10 = 40
Therefore, first four terms are: 10, 20, 30, 40
(ii) First term = a = –2 , d=0
Second term = a+d = –2 + 0 = –2
Third term = second term + d = –2 + 0 = –2
Fourth term = third term + d = –2 + 0 = –2
Therefore, first four terms are:–2, –2, –2, –2
(iii) First term = a = 4, d=–3
Second term = a+d = 4 – 3= 1
Third term = second term + d = 1 – 3= –2
Fourth term = third term + d = –2 – 3= –5
Therefore, first four terms are: 4, 1, –2, –5
(iv) First term = a = –1, d= ½
Second term = a+d = –1+ ½ =−½
Third term = second term + d = −½ + ½ = 0
Fourth term = third term + d = 0 + ½ = ½
Therefore, first four terms are:–1, −½, 0, ½
(v) First term = a =–1.25, d = –0.25
Second term = a+d = –1.25 – 0.25= –1.50
Third term = second term + d = –1.50 – 0.25 = –1.75
Fourth term = third term + d = –1.75 – 0.25 = –2.00
Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00

2. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Ans. Here a11=38 and a16=73
Using formula an=a+(n−1)d,to find nth term of arithmetic progression,
38=a+(11−1)(d) And 73=a+(16−1)(d)
⇒ 38=a+10And 73=a+15d
These are equations consisting of two variables.
We have, 38=a+10d
⇒ a=38−10d
Let us put value of a in equation (73=a+15d),
73=38−10d+15d
⇒ 35=5d
Therefore, Common difference =d=7
Putting value of d in equation 38=a+10d,
38=a+70
⇒ a=−32
Therefore, common difference = d = 7 and First term = a = –32
Using formula an=a+(n−1)d, to find nth term of arithmetic progression,
a31=−32+(31−1)(7)=−32+210=178
Therefore, 31stterm of AP is 178.

3. If the third and the ninth terms of an AP are 4 and –8 respectively, which term of this AP is zero?
Ans. It is given that 3rd and 9th term of AP are 4 and –8 respectively.
It means a3=4 and a9=−8
Using formula an=a+(n−1)d, to find nth term of arithmetic progression,
4 = a + (3 – 1)d And, –8 = a + (9 – 1)d
⇒ 4=a+2And, −8=a+8d
These are equations in two variables.
Using equation 4=a+2d, we can say that a=4−2d
Putting value of a in other equation −8=a+8d,
−8=4−2d+8d
⇒ −12=6d
⇒ d=−2
Putting value of d in equation −8=a+8d,
−8=a+8(−2)
⇒ −8=a–16
⇒ a=8
Therefore, first term =a=8 and Common Difference =d=−2
We want to know which term is equal to zero.
Using formula an=a+(n−1)d, to find nth term of arithmetic progression,
0=8+(n−1)(−2)
⇒ 0=8−2n+2
⇒ 0=10−2n
⇒ 2n=10
⇒ n=5
Therefore, 5thterm is equal to 0.

4. Two AP’s have the same common difference. The difference between their 100thterms is 100, what is the difference between their 1000thterms.
Ans. Let first term of 1stAP = a
Let first term of 2ndAP = a
It is given that their common difference is same.
Let their common difference be d.
It is given that difference between their 100thterms is 100.
Using formula an=a+(n−1)d, to find nth term of arithmetic progression,
a+(100−1)d– [a′+(100−1)d] =a+99da′−99d=100
⇒ aa′=100 … (1)
We want to find difference between their 1000thterms which means we want to calculate:
a+(1000−1)d– [a′+(1000−1)d] =a+999da′−999d=a– a
Putting equation (1) in the above equation,
a+(1000−1)d– [a′+(1000−1)d] =a+999da′+999d=aa′=100
Therefore, difference between their 1000th terms would be equal to 100.

5. How many three digit numbers are divisible by 7?
Ans. We have AP starting from 105 because it is the first three digit number divisible by 7.
AP will end at 994 because it is the last three digit number divisible by 7.
Therefore, we have AP of the form 105,112,119…, 994
Let 994 is the nth term of AP.
We need to find n here.
First term = a = 105, Common difference = d = 112 – 105= 7
Using formula an=a+(n−1)d, to find nth term of arithmetic progression,
994=105+(n−1)(7)
⇒ 994=105 + 7− 7
⇒ 896 = 7n
⇒ n=128
It means 994 is the 128th term of AP.
Therefore, there are 128 terms in AP.

6. 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?
Ans. Penalty for first day = Rs 200, Penalty for second day = Rs 250
Penalty for third day = Rs 300
It is given that penalty for each succeeding day is Rs 50 more than the preceding day.
It makes it an arithmetic progression because the difference between consecutive terms is constant.
We want to know how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
So, we have an AP of the form 200, 250, 300, 350 … 30 terms
First term = a = 200, Common difference = d = 50, n = 30
Applying formula, to find sum of n terms of AP, we get ⇒ Sn = 15(400+29 × 50)
⇒ Sn = 15(400+1450) = 27750
Therefore, penalty for 30 days is Rs. 27750.

7. 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 two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Ans. There are three sections of each class and it is given that the number of trees planted by any class is equal to class number.
The number of trees planted by class I = number of sections × 1 = 3 × 1 = 3
The number of trees planted by class II = number of sections × 2 = 3 × 2 = 6
The number of trees planted by class III = number of sections × 3 = 3 × 3 = 9
Therefore, we have sequence of the form 3, 6, 9 … 12 terms
To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 … 12 terms.
First term = a = 3, Common difference = d= 6 – 3= 3 and n = 12
Applying formula, to find sum of n terms of AP, we get 8. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.
Each step has a rise of m and a tread of m (see figure). Calculate the total volume of concrete required to build the terrace.
Ans. Volume of concrete required to build the first step, second step, third step, ……. (in m2) are    Total volume of concrete required =     = 750 m3

9. For what value of n are the nth term of the following two AP’s are same 13, 19, 25,…. and 69, 68, 67 …
Ans. nth term of 13, 19, 25 ,………… = nth term of 69, 68, 67,………
13+(n – 1) 6 = 69 + (n – 1) (-1)
Therefore, n = 9

10. Check whether 301 is a term of the list of numbers 5, 11, 17, 32,……..?
Ans.  So, 301 is not a term of the given list.

11. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Ans. Put the valve of d in eq. (i) 12. Find the sum of AP in Ans.   13. In an AP, find and a.
Ans.   14. Find for the AP in -9, -14, -19, -24…
Ans.  15. Find the sum to n term of the AP in 5, 2, -1, -4, -7……….
Ans.  16. Find the sum of first 24 terms of the list of no. whose nth term is given by Ans. Put    