Important Questions for CBSE Class 10 Maths Chapter 5 - Arithmetic Progressions 4 Mark Question -CoolGyan

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CBSE Class 10 Maths Chapter-5 Arithmetic Progressions Important Questions

CBSE Class 10 Maths Important Questions Chapter 5 – Arithmetic Progressions

4 Mark Questions

1. For the following APs, write the first term and the common difference.
(i) 3, 1, –1, –3 …
(ii) –5, –1, 3, 7…
(iii)
(iv) 0.6, 1.7, 2.8, 3.9…
Ans. (i) 3, 1, –1, –3…
First term = a= 3,
Common difference (d) = Second term – firstterm = Third term – second term and so on
Therefore, Common difference (d) = 1 – 3= –2
(ii) –5, –1, 3, 7…
First term = a= –5
Common difference (d) = Second term –Firstterm
= Third term – Second term and so on
Therefore, Common difference (d) = –1 – (–5)= –1+5 =4
(iii)
First term = a=
Common difference (d) = Second term –First term
= Third term –Second term and so on
Therefore, Common difference (d) =
(iv) 0.6, 1.7, 2.8, 3.9…
First term = a= 0.6
Common difference (d) = Second term –Firstterm
= Third term –Secondterm and so on
Therefore, Common difference (d) = 1.7−0.6=1.1

2. The 17^thterm of an AP exceeds its 10^thterm by 7. Find the common difference
Ans. (i) We need to show that a₁,a₂…a_n form an AP where a_n=3+4n
Let us calculate values of a₁,a₂,a₃… using a_n=3+4n
a₁=3+4(1)=3+4=7 a₂=3+4(2)=3+8=11
a₃=3+4(3)=3+12=15 a₄=3+4(4)=3+16=19
So, the sequence is of the form 7,11,15,19…
Let us check difference between consecutive terms of this sequence.
11 – 7= 4, 15 – 11= 4, 19 – 15= 4
Therefore, the difference between consecutive terms is constant which means terms a₁,a₂…a_n form an AP.
We have sequence 7,11,15,19…
First term = a =7 and Common difference = d = 4
Applying formula, to find sum of n terms of AP, we get

Therefore, sum of first 15 terms of AP is equal to 525.
(ii) We need to show that a₁,a₂…a_n form an AP where a_n=9−5n
Let us calculate values of a₁,a₂,a₃… using a_n=9−5n
a₁=9 – 5(1)=9 – 5=4 a₂=9 – 5(2)=9 – 10=−1
a₃=9 – 5(3)=9 – 15=−6 a₄=9 – 5(4)=9 – 20=−11
So, the sequence is of the form 4,−1,−6,−11…
Let us check difference between consecutive terms of this sequence.
–1–(4)= –5, –6–(–1) = –6 + 1 = –5, –11–(–6) = –11 + 6 = –5
Therefore, the difference between consecutive terms is constant which means terms a₁,a₂…a_n form an AP.
We have sequence 4,−1,−6,−11…
First term = a =4 and Common difference = d = –5
Applying formula, to find sum of n terms of AP , we get

Therefore, sum of first 15 terms of AP is equal to –465.

3. 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 than its preceding term, find the value of each of the prizes.
Ans. It is given that sum of seven cash prizes is equal to Rs 700.
And, each prize is R.s 20 less than its preceding term.
Let value of first prize = Rs. a
Let value of second prize =Rs (a−20)
Let value of third prize = Rs (a−40)
So, we have sequence of the form:
a, a−20, a−40, a – 60…
It is an arithmetic progression because the difference between consecutive terms is constant.
First term = a, Common difference = d = (a – 20) – a= –20
n = 7 (Because there are total of seven prizes)
S₇ = Rs 700 {given}
Applying formula, to find sum of n terms of AP, we get
⇒
⇒ 200=2a–120
⇒ 320=2a
⇒ a=160
Therefore, value of first prize = Rs 160
Value of second prize = 160 – 20= Rs 140
Value of third prize = 140 – 20= Rs 120
Value of fourth prize = 120 – 20 = Rs 100
Value of fifth prize = 100 – 20 = Rs 80
Value of sixth prize = 80 – 20 = Rs 60
Value of seventh prize = 60 – 20 = Rs 40

4. A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … What is the total length of such a spiral made up of thirteen consecutive semicircle.

Ans. Length of semi–circle =
Length of semicircle of radii 0.5 cm = (0.5) cm
Length of semicircle of radii 1.0 cm = (1.0) cm
Length of semicircle of radii 1.5 cm = (1.5) cm
Therefore, we have sequence of the form:
(0.5),(1.0),(1.5) … 13 terms {There are total of thirteen semicircles}.
To find total length of the spiral, we need to find sum of the sequence (0.5),(1.0),(1.5) … 13 terms
Total length of spiral =(0.5)+(1.0)+(1.5) … 13 terms
⇒ Total length of spiral = (0.5+1.0+1.5)… 13 terms … (1)
Sequence 0.5, 1.0, 1.5 …13 terms is an arithmetic progression.
Let us find the sum of this sequence.
First term = a= 0.5, Common difference = 1.0 – 0.5 = 0.5 and n = 13
Applying formula, to find sum of n terms of AP, we get

Therefore, 0.5 + 1.0 + 1.5 + 2.0 …13 terms = 45.5
Putting this in equation (1), we get
Total length of spiral = (0.5+1.5+2.0+ …13 terms) = (45.5) = 143 cm

5. 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. In how many rows are the 200 logs placed and how many logs are in the top row?
Ans. The number of logs in the bottom row = 20
The number of logs in the next row = 19
The number of logs in the next to next row = 18
Therefore, we have sequence of the form 20, 19, 18 …
First term = a = 20, Common difference = d = 19 – 20 = –1
We need to find that how many rows make total of 200 logs.
Applying formula, to find sum of n terms of AP, we get
⇒ 400=n(40 – n+1)
⇒ 400=40n−n²+n
⇒ n²−41n+400=0
It is a quadratic equation, we can factorize to solve the equation.
⇒ n²−25n−16n+400=0
⇒ n(n−25) – 16(n−25)=0
⇒ (n−25)(n−16)
⇒ n=25,16
We discard n = 25 because we cannot have more than 20 rows in the sequence. The sequence is of the form: 20, 19, 18 …
At most, we can have 20 or less number of rows.
Therefore, n = 16 which means 16 rows make total number of logs equal to 200.
We also need to find number of logs in the 16th row.
Applying formula, to find sum of n terms of AP, we get
200= 8 (20+l)
⇒ 200 = 160 + 8I
⇒ 40 = 8l
⇒ l= 5
Therefore, there are 5 logs in the top most row and there are total of 16 rows.

6. In a potato race, a bucket is placed at the starting point, which is 5 meters from the first potato, and the other potatoes are placed 3 meters apart in a straight line. There are ten potatoes in the line. 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?

Ans. The distance of first potato from the starting point = 5 meters
Therefore, the distance covered by competitor to pickup first potato and put it in bucket = 5 × 2 = 10 meters
The distance of Second potato from the starting point = 5 + 3 = 8 meters {All the potatoes are 3 meters apart from each other}
Therefore, the distance covered by competitor to pickup 2nd potato and put it in bucket = 8 × 2 = 16 meters
The distance of third potato from the starting point = 8 + 3 = 11 meters
Therefore, the distance covered by competitor to pickup 3rd potato and put it in bucket = 11 × 2 = 22 meters
Therefore, we have a sequence of the form 10, 16, 22…10 terms
(There are ten terms because there are ten potatoes)
To calculate the total distance covered by the competitor, we need to find :
10 + 16 + 22 + … 10 terms
First term = a = 10, Common difference = d =16 – 10= 6
n = 10 {There are total of 10 terms in the sequence}
Applying formula, to find sum of n terms of AP, we get

Therefore, total distance covered by competitor is equal to 370 meters.

7. Which term of the sequence is the first negative term?
Ans. For first negative term

28^th term is first negative term.

8. The p^th term of an AP is q and q^th term is p. Find its term.
Ans.

Put the value of d in eq (i)

9. If m times the m^th term of an A.P is equal to n times its n^th term, show that the term of the AP is zero.
Ans.

10. The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the AP.
Ans. (Given)

On solving equations (i) and (ii)

First three terms are -13,-8,-3.

11. If the sum of n terms of an AP is and its m^th term is 164, find the value of m.
Ans.
Put

12. If the sum of three numbers in AP, be 24 and their product is 440, find the numbers.
Ans. Let no. be a – d, a, a + d
(Given)

.

13. If are in AP, then prove that are in AP.
Ans. Given are in AP
Then
If are in AP then

From (i) and (ii),
Hence proved.

14. If be the sum of n, 2n and 3n terms respectively of an AP, prove that
Ans.

R.H.S

15. The ratio of the sums of m and n terms of an AP is , show that the ratio of the m^th and n^th term is
Ans.

16. If the sum of first p terms of an AP is the same as the sum of its first q terms, show that the sum of the first term is zero.
Ans.

17. For the A.P
Ans.

18. In an AP p^th, q^th and r^th terms are respectively a, b and c. Prove that
Ans.

Similarly,

Adding (iv), (v) and (vi)

19. If term of an A.P is twice the term, show that term is twice the term.
Ans.

20. The sum of four numbers in AP is 50 and the greatest number four times the least. Find the numbers.
Ans. Let no. be

21. Find the sum of all integers between 84 and 719 which are multiples of 5.
Ans.

22. If m^th term of an A.P is and the n^th term is show that the sum of terms is
Ans.

On solving (i) and (ii),

23. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 14th of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.
(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.
Ans. (i) Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15+8 = Rs 23
Taxi fare after 3 km = 23+8 = Rs 31
Taxi fare after 4 km = 31 +8 = Rs 39
Therefore, the sequence is 15, 23, 31, 39…
It is an arithmetic progression because difference between any two consecutive terms is equal which is 8.(23 – 15=8, 31 – 23=8, 39 – 31=8,…)
(ii) Let amount of air initially present in a cylinder = V
Amount of air left after pumping out air by vacuum pump =
Amount of air left when vacuum pump again pumps out air =
So, the sequence we get is like
Checking for difference between consecutive terms …

Difference between consecutive terms is not equal.
Therefore, it is not an arithmetic progression.
(iii) Cost of digging 1 meter of well = Rs 150
Cost of digging 2 meters of well = 150+50=Rs 200
Cost of digging 3 meters of well = 200+50 = Rs 250
Therefore, we get a sequence of the form 150, 200, 250…
It is an arithmetic progression because difference between any two consecutive terms is equal.(200 – 150=250 – 200= 50…)
Here, difference between any two consecutive terms which is also called common difference is equal to 50.
(iv) Amount in bank after Ist year =  … (1)
Amount in bank after two years =  … (2)
Amount in bank after three years =  … (3)
Amount in bank after four years =  … (4)
It is not an arithmetic progression because (2)−(1)≠(3)−(2)
(Difference between consecutive terms is not equal)
Therefore, it is not an Arithmetic Progression.

24. In the following AP’s find the missing terms:
(i) 2,__ , 26
(ii) __, 13, __, 3
(iii) 5, __, __,
(iv) –4. __, __, __, __, 6
(v) __, 38, __, __, __, –22
Ans. (i) 2, __ , 26
We know that difference between consecutive terms is equal in any A.P.
Let the missing term be x.
x– 2=26–x
⇒ 2x=28 ⇒ x=14
Therefore, missing term is 14.
(ii) __, 13, __, 3
Let missing terms be x and y.
The sequence becomes x, 13, y, 3
We know that difference between consecutive terms is constant in anyA.P.
y– 13=3–y
⇒ 2y=16
⇒ y=8
And 13 – x=y–13
⇒ x+y=26
But, we have y=8,
⇒ x+8=26
⇒ x=18
Therefore, missing terms are 18 and 8.
(iii) 5, __, __,
Here, first term = a=5 And, 4^thterm =a₄=
Using formula a_n=a+(n−1)d, to find n^thterm of arithmetic progression,
a₄=5+(4−1)d ⇒ =5+3d
⇒ 19= 2 (5 + 3d)
⇒ 19 = 10 + 6d
⇒ 6d = 19 – 10
⇒ 6d = 9 ⇒ d=
Therefore, we get common difference = d =
Second term = a+d =
Third term = second term + d =
Therefore, missing terms are and 8
(iv) –4. __, __, __, __, 6
Here, First term = a = –4 and 6^thterm = a₆ = 6
Using formula a_n=a+(n−1)d, to find n^th term of arithmetic progression,
a₆=−4+(6−1)d ⇒ 6=−4+5d
⇒ 5d=10 ⇒ d=2
Therefore, common difference = d = 2
Second term = first term + d = a+ d = –4 + 2 = –2
Third term = second term + d = –2 + 2 = 0
Fourth term = third term + d = 0 + 2 = 2
Fifth term = fourth term + d = 2 + 2 = 4
Therefore, missing terms are –2, 0, 2 and 4.
(v) __, 38, __, __, __, –22
We are given 2^ndand 6^thterm.
Using formula a_n=a+(n−1)d, to find n^th term of arithmetic progression,
a₂=a+(2−1)d And a₆=a+(6−1)d
⇒ 38=a+d And −22=a+5d
These are equations in two variables, we can solve them using any method.
Using equation (38=a+d), we can say that a=38−d.
Putting value of a in equation (−22=a+5d),
−22=38 – d+5d
⇒ 4d=−60
⇒ d=−15
Using this value of d and putting this in equation 38=a+d,
38=a–15
⇒ a=53
Therefore, we get a=53 and d=−15
First term = a = 53
Third term = second term + d = 38 – 15= 23
Fourth term = third term + d = 23 – 15= 8
Fifth term = fourth term + d = 8 – 15= –7
Therefore, missing terms are 53, 23, 8 and –7.

25. The given figure depicts a racing track whose left and right ends are semi-circular. The difference between the two inner parallel line segments is 60m and they are each 106 m long. If the track is 10m wide, find:
(i) The distance around the track along its inner edge,
(ii) The area of the track

Ans. (i) The distance around the track along the inner edge

(ii) The area of the track

26. Which of the following are APs? If they form an AP, find the common difference d and write threemore terms.
(i) 2,4,8,16…
(ii) 2,, 3, . ..
(iii) −1.2,−3.2,−5.2,−7.2…
(iv) −10,−6,−2,2…
(v)
(vi) 0.2,0.22,0.222,0.2222…
(vii) 0,−4,−8,−12…
(viii)
(ix) 1,3,9,27…
(x) a,2a,3a,4a…
(xi) a,a²,a³,a⁴…
(xii)
(xiii)
(xiv) 1²,3²,5²,7²…
(xv) 1²,5²,7²,73…
Ans. (i) 2,4,8,16…
It is not an AP because difference between consecutive terms is not equal.
As 4 – 2≠8−4
(ii) 2,, 3, …
It is an AP because difference between consecutive terms is equal.

Common difference (d) = ½
Fifth term =  Sixth term = 4+ ½ =
Seventh term =
Therefore, next three terms are 4,and 5.
(iii) −1.2,−3.2,−5.2,−7.2…
It is an AP because difference between consecutive terms is equal.
−3.2−(−1.2)=−5.2−(−3.2)=−7.2−(−5.2)=−2
Common difference (d) = −2
Fifth term = −7.2 – 2=−9.2
Sixth term = −9.2 – 2=−11.2
Seventh term = −11.2 – 2=−13.2
Therefore, next three terms are −9.2,−11.2 and −13.2
(iv) −10,−6,−2,2…
It is an AP because difference between consecutive terms is equal.
−6−(−10) =−2−(−6) =2−(−2)=4
Common difference (d) = 4
Fifth term = 2 + 4 = 6 Sixth term = 6 + 4 = 10
Seventh term = 10 + 4 = 14
Therefore, next three terms are 6,10 and 14
(v)
It is an AP because difference between consecutive terms is equal.

Common difference (d) =
Fifth term =
Sixth term =
Seventh term =
Therefore, next three terms are
(vi) 0.2,0.22,0.222,0.2222…
It is not an AP because difference between consecutive terms is not equal.
0.22−0.2≠0.222−0.22
(vii) 0,−4,−8,−12…
It is an AP because difference between consecutive terms is equal.
−4 – 0=−8−(−4)=−12−(−8)=−4
Common difference (d) = −4
Fifth term = −12 – 4=−16 Sixth term = −16 – 4=−20
Seventh term = −20 – 4=−24
Therefore, next three terms are −16, −20 and −24
(viii)
It is an AP because difference between consecutive terms is equal.

Common difference (d) = 0
Fifth term =  Sixth term =
Seventh term =
Therefore, next three terms are
(ix) 1,3,9,27…
It is not an AP because difference between consecutive terms is not equal.
3 – 1≠9−3
(x) a,2a,3a,4a…
It is an AP because difference between consecutive terms is equal.
2a– a=3a−2a=4a−3a=a
Common difference (d) = a
Fifth term = 4a+a=5a Sixth term = 5a+a=6a
Seventh term = 6a+a=7a
Therefore, next three terms are 5a,6a and 7a
(xi) a,a²,a³,a⁴…
It is not an AP because difference between consecutive terms is not equal.
a² – a≠a³−a²
(xii)
It is an AP because difference between consecutive terms is equal.

Common difference (d) =
Fifth term =  Sixth term =
Seventh term =
Therefore, next three terms are
(xiii)
It is not an AP because difference between consecutive terms is not equal.

(xiv) 1²,3²,5²,7²…
It is not an AP because difference between consecutive terms is not equal.
3²−1²≠5²−3²
(xv) 1²,5²,7²,73…  1,25,49,73…
It is an AP because difference between consecutive terms is equal.
5²−1²=7²−5²=73−7²=24
Common difference (d) = 24
Fifth term = 73+24=97 Sixth term = 97+24=121
Seventh term = 121 + 24 = 145
Therefore, next three terms are 97,121 and 145