HOTS & Value Based Questions: Arithmetic Progressions

# Class 10 Maths Chapter 5 HOTS Questions - Arithmetic Progressions

## HOTS Questions

Q1: Find the ‘6th’ term of the A.P. :
Sol:  Here,

∴ d = a2 – a1

Now, an = a + (n – 1)d

Thus, the nth term is

Again, we have

i.e., the 6th term is

Q2:  If the ratio of the sum of first n terms of two A.P.'s is (7n + 1) : (4n + 27), find the ratio of their mth terms.
Sol:  Let the first terms of given AP's be a1 and a2, common differences be d1 and d2 and let Sn and the sum of n terms.

Now, replacing n by (2m-1) for getting ratio of mth terms of given APs,

Thus, the required ratio of mth term of given AP's is (14m – 6) : (8m + 23)

Q3:  If the numbers a, b, c, d and e form an A.P., then find the value of a – 4b + 6c – 4d + e
Sol: We have the first term of A.P. as ‘a’.
Let D be the common difference of the given A.P.,
Then :
b = a + D,  c = a + 2D, d= a + 3D and e = a + 4D

∴ a – 4b + 6c – 4d + e
=   a – 4(a + D ) + 6 (a + 2D) – 4 (a + 3D) + (a + 4D)
= a – 4a + 6a – 4a + a – 4D + 12D – 12D + 4D
= 8a – 8a + 16D – 16D = 0
Thus, a – 4b + 6c – 4d + e = 0

Q4:is the arithmetic mean between ‘a’ and ‘b’, then, find the value of ‘n’.
Sol:  Note: A.M., between ‘a’ and ‘b ’= 1/2
(a + b)
We know that :
A.M. between ‘a’ and ‘b’ = a+b/2
It is given that,
is the A.M. between ‘a’ and ‘b’
∴

By cross multiplication, we get :

⇒2an + 1 + 2bn + 1 = an + 1 + abn + anb + bn + 1
⇒ 2an + 1 – an + 1 + 2bn + 1– bn + 1= abn + anb
⇒ an + 1 + bn + 1 = abn +anb
⇒ an+1 – anb = abn – bn+1
⇒ an[a–b] = bn[a –b]

⇒ n =  0

Q5:  Solve the equation :
1 + 4 + 7 + 10 + ... + x = 287
Sol:

Since,
∴ a = 1,d = 3 and an = x
∴ an = a + (n – 1)d
⇒ x = 1 + (n – 1) 3   or   x = 3n – 2

Also, Sn = n/2 (a+l)

⇒ 2(287) = n[1 + (3n – 2)]
⇒ 574 = n[3n – 1]
⇒ 3n2 – n – 574 = 0
Solving the above quadratic equation, we get

or

But, negative n is not desirable.
∴ n = 14
x = 3n – 2
Now, x = 3(14) – 2 = 42 – 2 = 40
Thus, x = 40

Q6: Find three numbers in A.P. whose sum is 21 and their product is 231.
Sol: Let the three numbers in A.P. are:
a – d,    a,   a + d
∴ (a – d) + a + (a + d) = 21
⇒ a – d + a + a + d = 21
or 3a = 21 ⇒   a = 7
Also, (a – d) × a × (a + d) = 231
∴ (7 – d) × 7 × (7 + d) = 231
⇒ (7 – d) (7 + d) × 7 = 231
⇒ 72 – d2 = 231 /7 = 33
⇒ 49 – d2 = 33
or d2 = 49 – 33 = 16
⇒ d = ± 4
Now, when d = 4, then three numbers in AP are : (7 – 4), 7,  (7 + 4) i.e. 3, 7, 11.
When d = –4, then three numbers in AP are : [7 – (–4)],  7,  [7 + (–4)]
or  11, 7, 3

Q7: The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and common difference.
Sol:  Let 'a' be the first term and 'd' be the common difference
Since, an = a + (n–1)d
∴ a9 = a + 8d  and a12 = a + 11d
Also a2 = a + d   and a3 = a + 2d
since a9 = 7a2   and a12 = 5a3 + 2
or a9 = 7 (a + d)  ⇒ a9  = 7a + 7d     ... (1)
a12 = 5 (a + 2d) + 2
⇒ a12 = 5a + 10d + 2           ... (2)

Now, a + 8d = 7a + 7d            [From (1)]
or – 6a +   d = 0  ... (3)
Also a + 11d = 5a + 10d + 2    [From (2)]
or –4a + d = 2         ... (4)

Subtracting (4) from (3), we have
–2a = –2   ⇒  a = 1
Now, from (3), –6 + d = 0  ⇒  d = 6
Thus, a = 1   and  d = 6

## Value-Based Questions

Q1: Savita has two options to buy a house:
(a) She can pay a lumpsum amount of ₹ 22,00,000
Or
(b) She can pay 4,00,000 cash and balance in 18 annual instalments of ₹ 1,00,000 plus 10% interest on the unpaid amount.
She prefers option
(i) and donates 50% of the difference of the costs in the above two options to the Prime Minister Relief Fund.
(i) What amount was donated to Prime Minister Relief Fund?
(ii) Which mathematical concept is used in the above problem?
(iii) By choosing to pay a lumpsum amount and donating 50% of the difference to the Prime Minister Relief Fund, which value is depicted by Savita?
Sol:
(a) Total cost of the house = ₹ 22,00,000
(b) Cash payment = ₹ 4,00,000
Balance = ₹ 22,00,000 – ₹ 4,00,000 = ₹ 18,00,000
1st instalment = ₹ [1,00,000 + 10% of balance]

= ₹ [1,00,000 + 1,80,000]   = ₹ 2,80,000
Balance after 1st instalment = ₹ [18,00,000 – 1,00,000]  = ₹ 17,00,000

2nd instalment = ₹ [1,00,000 + 10% of 17,00,000]
= ₹ [1,00,000 + 1,70,000]  = ₹ [2,70,000]
Balance after 2nd instalment = ₹ 17,00,000 – ₹ 1,00,000  = ₹ 16,00,000

∵ 3rd instalment = ₹ [1,00,000 + 10% of 16,00,000]
= ₹ [1,00,000 + 1,60,000]  = ₹ 2,60,000
... and so on.
∵ Total amount in instalments = ₹ 2,80,000 + ₹ 2,70,000 + ₹ 2,60,000 + ..... to 18 terms
where a = 2,80,000,  d = – 10,000, n = 18

= ₹ 9 [560,000+17(-10,000)
= ₹ 9 [560,000 - 170,000]
= ₹ 9 = [390,000] = ₹ 35,10,000
∴ Total cost of house = ₹ 35,10,000 + 4,00,000  = ₹ 39,10,000
Difference in costs of the house in two options
= ₹ 39,10,000 – ₹ 22,00,000  = ₹ 17,10,000
∴ (i) Amount donated towards Prime Minister Relief Fund = 50% of ₹ 17,10,000

(ii) Arithmetic Progressions
(iii) National Loyalty

Q2: In a school, students decided to plant 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 double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. Which value is shown in this question?
Sol: ∴ There are 12 classes in all.
Each class has 2 sections.
∴ Number of plants planted by class I = 1 x 2 = 2
Number of plants planted by class II = 2 x 2 = 4
Number of plants planted by class III = 3 x 2 = 6
Number of plants planted by class IV = 4 x 2 = 8
......................................................................................................
......................................................................................................
Number of plants planted by class XII = 12 x 2 = 24
The numbers 2, 4, 6, 8, ........................ 24 forms an A.P.
Here, a = 2, d = 4 – 2 = 2
∵ Number of classes = 12
∴ Number of terms (n) = 12

Now, the sum of n terms of the above A.P., is given by Sn = n/2 [2a+(n-1)d]
∴ S12 = 12/2 [2(2) -(12)-1) 2]
= 6 [4 + (11 x 2)]
= 6 x 26 = 156
Thus, the total number of trees planted = 156
Value shown: To enrich polution free environment.

The document Class 10 Maths Chapter 5 HOTS Questions - Arithmetic Progressions is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 5 HOTS Questions - Arithmetic Progressions

 1. What are arithmetic progressions in mathematics?
Ans. Arithmetic progressions are a sequence of numbers in which the difference between any two consecutive terms is constant. For example, 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.
 2. How can arithmetic progressions be used in real-life situations?
Ans. Arithmetic progressions can be used in various real-life situations such as calculating interest rates, predicting population growth, and analyzing financial data. For example, if a person saves \$100 every month, the total amount saved after a certain number of months can be calculated using an arithmetic progression formula.
 3. How can we find the nth term of an arithmetic progression?
Ans. To find the nth term of an arithmetic progression, we use the formula: nth term = first term + (n - 1) * common difference. By substituting the values of the first term, common difference, and the desired value of n, we can easily calculate the nth term.
 4. What is the sum of the first n terms of an arithmetic progression?
Ans. The sum of the first n terms of an arithmetic progression can be calculated using the formula: Sum = (n/2) * (2 * first term + (n - 1) * common difference). By substituting the values of n, first term, and common difference, we can find the sum of the desired number of terms.
 5. How can arithmetic progressions be used to solve problems involving time and distance?
Ans. Arithmetic progressions can be used to solve problems involving time and distance, such as calculating the average speed of a moving object. By considering the time intervals and respective distances traveled, we can form an arithmetic progression to find the average speed.

## Mathematics (Maths) Class 10

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