Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions

Ans:
Let the AP be a, a + d, a + 2d,.............


⇒ 3a + 27d= a + 29d
⇒ 2a – 2d = 0
⇒ 2a + 2d
∴ a = d ...(i)     
Now,  S₆ = 42

⇒ 3[2a + 5d] = 42
⇒ 2a + 5d =14
⇒ 2a + 5a = 14 [From eqn (i)]
⇒ 7a = 14
⇒ a = 14/7
∴ a = 2
So First term = 2
Common difference = 2.

Ans:
S₇ = 49

⇒
⇒ 2a + 6d = 14
⇒ a + 3d = 7 ...(i)
S₁₇ = 289


⇒ 2(a + 8d) = 34
⇒ a + 8d = 17 ...(ii)  
From eqn (i) and (ii).    
–5d = –10
∴  d =2  
Put the value of d in eqn.(i),
∴ a + 3 × 2 =7
⇒ a + 6 = 7
⇒ a = 7 - 6
⇒ a = 1
∴

Ans: (c)
Sol: We have, a, b, c are in A.P.
b = a + d, and c =  a + 2d
Now, a - 2b - c = a - 2(a + d) -  (a + 2d)
= a - 2a - 2d - a - 2d
= -2a- 4d

Ans: (a)
Sol: Since, k + 2, 4k - 6 and 3k - 2 are three consecutive terms of A.P.
a₂ - a₁ = a₃ - a₂
⇒ (4k - 6)- (k + 2) = (3k - 2) - (4k - 6)
⇒ 4k -6 - k - 2= 3k - 2 - 4k + 6
⇒ 3k - 8 = -k + 4
⇒ 4k = 12
⇒ k = 3

Ans: We have
First term, a₁ = - 14
Fifth term, a₅ = 2
Last term, a_n = 62
Let d be the common difference and n be the number of terms.
∵ a₅ = 2
⇒ -14 +(5 - 1)d = 2
⇒  4d = 16
⇒  d =4
Now, a_n = 62
⇒  -14 + (n - 1)4 = 62
⇒ 4n - 4 = 76
⇒ 4n = 80
⇒ n= 20
There are 20 terms in A.P.

Ans: Given, A.P. is 65, 61, 57, 53,.....
Here, first term a = 65 and common difference, d = -4
Let the n^th term is negative.
Last term, a_n = a + (n - 1) = 65 + (n - 1)(-4)
= 65 - 4n + 4
= 69 - 4n, which will be negative when n = 18
So, 18^th term is the first negative term.

Ans: (b)
Sol: Since, a, b ,c are in A.P. then b - a = c -b
⇒ 2b = a + c
First n odd natural number be 1, 3, 5 ..... (2n - 1).
which form an A.P. with a = 1 and d = 2
Sum of first n odd natural number = n/2[2a + (n -1)d]
= n/2 [2 + (n - 1)2] = n²
Hence, assertion and reason are true but reason is not the correct expiation of assertion.

Ans: Here, a = 15 and S₁₅ = 750
∵ S_n = n/2[2a + (n -1)d]
∴ S₁₅ = 15/2 [2 x 15 + (15 -1)d] = 750
⇒ 15(15 + 7d) = 750
⇒ 15 + 7d = 50
⇒ 7d = 35
⇒ d = 5
Now, 20^th term = a + (n - 1)d
= 15 + (20 - 1) 5
= 15 + 95
= 110

Ans: Total amount of loan Rohan takes = Rs. 1,18,000
First instalment paid by Rohan = Rs. 1000
Second instalment paid by Rohan = 1000 + 100 = Rs. 1100
Third instalment paid by Rohan = 1100 + 100 = Rs. 1200 and so on.
Let its 30^th instalment be n.
Thus, we have 1000,1100,1200, which forms an A.P. with first term (a) = 1000
and common difference (d) = 1100 - 1000 = 100
n^th term of an A.P. a_n= a + (n - 1)d
For 30^th instalment, a₃₀ = a + (30 - 1)d
= 1000 + (29) 100 = 1000 + 2900 = 3900
So Rs. 3900 will be paid by Rohan in the 30^th instalment.
Now, we have a = 1000, last term (l)= 3900
Sum of 30 instalment, S₃₀ = 30/2[a + 1]
⇒ S₃₀ = 15(1000 + 3900) = Rs. 73500
Total amount he still have to pay after the 30th instalment = (Amount of loan) - (Sum of 30 instalments)
= Rs. 1,18,000 - Rs. 73,500 = Rs. 44,500
Hence, Rs. 44,500 still have to pay after the 30^th instalment.

Ans: Let a and d be the first term and common difference of an AP.
Given that, a₁₁ : a₁₈ = 2 : 3

⇒ 3a + 30d = 2a + 34d
⇒ a = 4d ...(i)
Now, a₅ = a + 4d = 4d + 4d = 8d [from Eq.(i)]
And a₂₁ = a + 20d = 4d + 20d = 24d [from Eq. (i)]
a₅ : a₂₁ = 8d : 24d = 1 : 3
Now, sum of the first five terms, S₅ = 5/2 [2a + (5−1)d]
= 5/2 [2(4d) + 4d]   [from Eq.(i)]
= 5/2 (8d + 4d) = 5/2 × 12d = 30d
And, sum of the first 21 terms, S₂₁ = 21/2 [2a + (21−1)d]
= 21/2 [2(4d) + 20d]= 21/2 × 28 d = 294 d from Eq..(i)]
So, ratio of the sum of the first five terms to the sum of the first 21 terms is,
S₅ : S₂₁ = 30d : 294d = 5 : 49

Ans: Let there be n rows to pile of 250 logs
Here, the bottom row has 22 logs and in next row, 1 log reduces
It means, we get an AP 22, 21, 20, 19, ..................... n with first term or a = 22 and d = -1
Now, we know that total logs are 250 or we can say that,
S_n =250
Since sum of n terms of an A.P. S_n = n/2 (2a + (n-1) d)
= 250 Therefore, n/2 (2 x 22 + (n-1) x (-1))
or 500 = n (44 - (n-1))
500 = n (45- n)
n² - 45 n + 500 = 0
By solving this, we get (n-20) (n-25) = 0
Since, there are 22 logs in first row and in next row, 1 log reduces, then we can not have more than 22 terms
∴ n ≠ 25
and n = 20
Means, 20^th row is the top row of the pile
Now let's find out number of logs in 20^th row
We know that value of nth term of an A.P. = a + (n-1) d
N₂₀ = [22 + (20-1) (-1)]
=(22-19) = 3
Therefore, there are 3 logs in the top row.

Ans: (d)
To find the next term of the arithmetic progression (A.P.) √$\backslash sqrt\{7\},\; \backslash sqrt\{28\},\; \backslash sqrt\{63\}$7, √28, √63, let's first determine the common difference.
We need the approximate values for calculation:

$\mathrm{\surd 7 }$≈ 2.6458

$\mathrm{\surd 28}$ ≈ 5.2915

$\mathrm{\surd 63}$ ≈ 7.9373

So:
d = $\mathrm{\surd }$28 − $\mathrm{\surd }$7 ≈ 5.2915 - 2.6458 = 2.6457
Similarly, checking the difference between $\mathrm{\surd }$63 and$\mathrm{ \surd }$$\backslash sqrt\{28\}$28:
d = $\mathrm{\surd }$63 - $\mathrm{\surd }$28 ≈ 7.9373 − 5.2915 = 2.6458
The common difference $dd$d is approximately 2.6458, so the sequence is indeed an A.P.
The next term after $\mathrm{\surd }$63 is:
Next term = $\mathrm{\surd }$63 + d ≈ 7.9373 + 2.6458 = 10.5831
Now, approximate this result as the square root of the next perfect square:
10.5831 ≈ 112
Thus, the next term is: √$\backslash sqrt\{112\}$112

Ans: Since a, 7, b,  23 are in A.P.
∴ Common difference is same.
∴ 7 - a = b -7 = 23 - b
Taking second and third terms, we get
b - 7 = 23 - b
⇒ 2b = 30
⇒ b = 15
Taking first and second terms, we get
⇒ 7 - a = b - 7
⇒ 7 - a = 15 - 7
⇒ 7 - a = 8
⇒ a = -1
Hence, a = -1, b = 15.

Ans: Given, 293, 285, 277..... 53 be an A.P.
a = 293, d = 285 - 293 = -8
We know. a_n = a + (n - 1 )d
⇒ 53 = 293 =  (n - 1)(-8)
⇒ 53 - 293 = (n - 1) (-8)
⇒ -240 =  (n - 1) (-8)
⇒ 30 = n - 1
⇒ n = 31

Ans: Let the first term and common difference of an A.P. be a and d, respectively.
Given a₃= 5 and a₇ = 9
a + (3 - 1 ) d = 5 and a + (7 - 1)d  =  9
a + 2d = 5 --------------(i)
and a + 6d = 9--------------(ii)
On subtracting (i) from (ii), we get
⇒ 4d  = 4
⇒ d = 1
From (i), a + 2(1) = 5 ⇒ a + 2 = 5 ⇒ a = 3
So. required A.P. is a, a + d, a + 2d, a + 3d......
i.e. 3, 3 +1, 3 + 2(1), 3 + 3(1), .....,  i.e.. 3,  4,  5, 6, .....

Ans: (b)
Sol: Given, -5/7, a, 2 are in A.P. therefore common difference is same.
∴ a₂ - a₁ = a₃ - a₂

Ans: (c)
Sol: In option (c), We have

As a₂ - a_{1 ≠}  a₃- a₂ the given list of numbers does not form an A.P.

Ans: (a)
Sol: Given, 2x, (x + 10) and (3x + 2) are in A.P.
(x + 10) - 2x = (3x + 2) -(x + 10)
⇒ -x + 10= 2x - 8
⇒ - 3x = -18
⇒ x = 6

Ans: Let a₁ = (a - b)², a₂ = (a² + b²) and a₃= (a + b)²
Now. a₂ - a₁ = (a² + b²) - (a - b)²
= a² + b² - (a² + b² - 2ab)
= a² + b²- a²- b² + 2ab = 2ab
Again a₃ - a₂ = (a + b)² - (a² + b²)
= a² + b² + 2ab - a² - b² = 2ab
∴ a₂ - a₁ = a₃ - a₂
So, (a - b)², (a² + b²) and (a + b)² are in A.P.

Ans: Let the four consecutive numbers be (a - 3d), (a - d), (a + d), (a + 3d).

Sum of four numbers = 32 (Given)
⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32 ⇒  a = 8

If d = 2. then the numbers are (8 - 6), (8 - 2), (8 + 2) and (8 + 6) i.e., 2,6, 10, 14 .
If d = -2. then the numbers are (8 + 6), (8 + 2), (8 - 2). (8 - 6) i.e.,14 ,10 ,6 ,2 .
Hence, the numbers are 2, 6, 10, 14 or 14, 10, 6, 2.

Ans: First 100 natural numbers are 1 , 2 , 3 ...... 100 which form an A.P. with a = 1, d = 1.
Sum of n terms =S_n = n/2 [2a + (n - 1)d] 

 = 100/2 [2 x 1 + (100 - 1) x 1] = 50 [2 + 99] = 50 x 101 = 5050

Ans: Given, a₄ = -15 and a₉ = -30
a + 3d = - 15  (i)
a + 8d = -30   (ii)
On subtracting (ii)from (i), we have
-5d = 15
⇒ d = - 3
Put d = - 3 in (i), we have
a + 3(-3)= - 15 
⇒ a - 9 = - 15
⇒ a = - 6
Now, S_n = n/2 [2a + (n - 1)d]
⇒ S₁₆ = 16/2 [2(-6) + (16 - 1) (-3)]
= 8 [2(-6) + (15) (-3)] = 8 [-12 - 45] = -456

Ans:  Given, d = - 3, a = 54 and a_n= 0
Since a_n = a + (n -1)d
∴ 0 = 54 + (n - 1)(-3)
⇒ 0 = 54 - 3n + 3
⇒  3n = 57
⇒ n = 19
Now,
S_n = n/2 [2a+(n-1)d]
= 19/2 [2 × 54 +(19 - 1)(-3)]
= 19/2 [108 - 54] = 19/2 × 54 = 513

Ans: (−5) + (−8)+ (−11) + ... + (−230) .
Common difference of the A.P. (d) = a₂ - a₁ 
= -8-(-5)
= -8+5
= -3
So here,
First term (a) = −5
Last term (l) = −230
Common difference (d) = −3
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n-1) d
So, for the last term,
- 230 = -5 + ( n-1) (-3)
- 230 = -5-3n + 3
-23 +2 = -3n
- 228/-3 = n
n = 76
Now, using the formula for the sum of n terms, we get
S_n = 76/2 [2(-5) + (76-1)(-3)]
= 38 [-10 + (75)(-3)]
=38 (-10-225)
= 38(-235)
= -8930
Therefore, the sum of the A.P is  Sn = -8930 

Ans: Given: first term a₁ = a, second term, a₂ = b and last term, l = c.  
So, common difference, d = a₂ – a₁ = b – a  
Let this A.P. contains n terms.          
Then,  l = a + (n – 1)d 
⇒ c = a + (n – 1) (b – a) 
⇒ c – a = (n – 1)(b – a)

Now, sum of n terms of A.P. is given by,

=
⇒
Hence, proved.

Ans: Give A.P. is
√3, √12, √27, √48, ......
or √3, 2√3, 3√3,4√3, .......
∴ d = common difference = 2√3 - √3 = √3

Ans:  Let n^th term of A.P. 10, 7, 4, ...is -41
∴ a_n = a + (n - 1)d
⇒ - 41 = 10+(n-1)(-3)     [∵ d = 7 - 10 = -3]
⇒ - 41 = 10 - 3n + 3
⇒ -41 = 13 - 3n
⇒ 3n = 54=n= 18
∴ 18^th term of given A.P. is - 41 

Ans: Given, a = 15,  d = - 3 and a_n = 0
∴ a_n = a + (n - 1)d
⇒ 15 + (n - 1 )(- 3)= 0
⇒ 15  - 3n +3 = 0 
⇒ 18 - 3n = 0 
⇒ - 3n = -18
⇒ n = 6

Ans: Two-digit numbers which are divisible by 3 are 12, 15, 13..... 99. which forms an A.P. with first term (a) = 12, common difference (d) = 15 - 12 = 3 and last term (l) or n^th term = 99 
a + (n - 1)d = 99
⇒ 12 + (n - 1)3 = 99 
⇒ 3n = 99 - 9
⇒ n = 90/3
⇒ n = 30

Ans:  Given, a₉ = 0 . we have to show that a_{29 =} 2a₁₉
a + 8d = 0
⇒ a = - 8 d
Now, a_{19 =} a + 18d = -8d + 18d = 10d
a₂₉ = a + 28d = -8d + 28d = 20d = 2(10d ) = 2a₁₉
Hence, a_{29 =} 2a₁₉

Ans:  We have, first term, a = 3, common difference,  d = 15 - 3 =12
n^th term of an A.P. is given by a_n = a + (n - 1)d 
∴ a₂₁ =  3 + (20) x 12 
= 3 + 240 
= 243 
Let the r^th term of the AP. be 120 more than the 21^st term.
⇒ a + (r - 1) d = 243 + 120
⇒ 3 + (r - 1) 12 = 363
⇒ (r - 1) 12 = 360 ⇒ r - 1= 30 ⇒ r = 31

Ans:  According to question, a₁₇ - a₁₀ = 7 
i.e. a + 16d- (a + 9d) = 7
where a = first term d = common difference
⇒ 7d = 7
∴ d = 1

Ans: The saving in first month is₹ 150. 
The saving in second month is 
₹(150 + 50) = ₹ 200 
Similarly, saving goes on increasing every month by ₹ 50.   
Savings = ₹ 150, ₹ 200, ₹ 250, ₹300,..... 
Savings forms an A.P. in which first term (a) = 150 and common difference, (d) = 50 
Then, total savings for 12 months 

Then, Ramkali would be able to save ₹ 5,100 in 12 months and she needs ₹5,000 to send her daughter to school. 
Hence, Ramkali would be able to send her daughter to school. 
Values: Putting efforts to send her daughter to school shows her awareness regarding girls education and educating a child.

Ans: Let the value of first most expensive prize be ₹ a. 
Then, according to the given condition, prizes are a, a – 50, a – 100, a – 150 ....... 
The given series forms an A.P., with a common difference of (– 50). 
Here, first term = a 
Common difference    d = – 50 
Number of terms, n    = 10 and, 
sum of 10 terms, S₁₀ = ₹ 4,250 
By formula,

Hence, the value of the prizes are: ₹ 650, ₹ 600, ₹ 550, ₹ 500, ₹ 450, ₹ 400, ₹ 350, ₹ 300, ₹ 250, ₹ 200.

Ans: Since, child puts ₹ 5 on 1st day, ₹ 10 (2 × 5) on 
2^nd day, ₹ 15(3 × 5) on 3^rd day and so on.  
Total savings = 190 coins = 190 × 5 = ₹ 950          
So, the series of her daily savings is, ₹ 5, ₹ 10, ₹ 15, .....         
Clearly, this series is an A.P. 
So, first term, a = 5               
Common difference, d = 5               
Sum of total savings, S_n = 950 
Let, n be the last day when piggy bank becomes full.

n(n + 20) – 19(n + 20) = 0 
(n – 19) (n + 20) = 0 
n = 19 or – 20          
But ‘n‘ cannot be negative, hence n = 19.          
Hence, she continuous the savings for 19 days and saves ₹ 950.          
Views on habit of saving: 
(1) Child is developing a very good habit of savings.
(2) Consistent saving can create a wonder.

Ans: Given, A.P. is
Here, first term, a = 1 / m
Common difference,

∴ n^th term, a_n= a + (n – 1)d

Hence, the n^th term of given A.P. is .

Ans: It is given that, 
Money saved in 1^st month =₹ 32
Money saved in 2^nd month = ₹ 36 
Money saved in 3^rd month =₹ 40 
Let Yasmeen saves ₹ 2000 in n months. 
Clearly, monthly savings of Yasmeen are in A.P. with a = ₹ 32, d = ₹ 4 and S_n = ₹ 2000. 
We know that sum of first n terms of an A.P. is

Splitting the middle term, we get n² + 40n – 25n – 1000 = 0 
n(n + 40) – 25(n + 40) = 0 
(n + 40)(n – 25)  =  0 
⇒ n =  –40, n = 25 
But n ≠ –40, as number of months cannot be negative. 
∴ n = 25 Hence, 
Yasmeen will save ₹ 2000 in 25 months

Ans: Let the total time taken to catch the thief be ‘n’ minutes. 
Speed of thief    = 100 m/min., 
Total distance covered by the thief = 100n 
Now, the speed of the policeman in the first minute is 100 m/min, in the 2nd minute is 110 m/min, in the 3rd minute is 120 m/min and so on.
Then, the speed forms an A.P. with a constant increasing speed of 10 m/min, thus, the series is, 
100, 110, 120, 130, ... 
As the policeman starts after a minute, so time taken by the policeman to catch the thief is (n – 1) minutes. 
∴ Total distance covered by the policeman 
= 100 + 110 + 120 + ... (n – 1) terms

⇒ 10n² – 30n – 180 = 0 
⇒ n² – 3n – 18 = 0 
⇒ (n – 6) (n + 3)    = 0 
⇒ n = 6     [n = – 3, is not possible] 
Hence, the policeman takes 6 minutes to catch the thief.