Math Olympiad Test: Arithmetic Progression- 4 - Class 10 MCQ

# Math Olympiad Test: Arithmetic Progression- 4 - Class 10 MCQ

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## 10 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Arithmetic Progression- 4

Math Olympiad Test: Arithmetic Progression- 4 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Arithmetic Progression- 4 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Arithmetic Progression- 4 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Arithmetic Progression- 4 below.
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Math Olympiad Test: Arithmetic Progression- 4 - Question 1

### Deepak repays his total loan of ₹1,18,000 by paying every month starting with the first instalment of ₹1000. If he increases the instalment by ₹100 every month. what amount will be paid as the last instalment of loan?

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 1

1st instalment = ₹ 1000
2nd instalment = ₹ 1000 + ₹ 100 = ₹ 1100
3rd instalment = ₹ 1100 + ₹ 100 = ₹ 1200 and so on
Let number of instalments = n
∴ 1000 + 1100 + 1200 + ... up to n terms = 118000
⇒ n/2[2 x 1000 + (n - 1)100] = 118000
⇒ 100n2 + 1900n – 236000 = 0
⇒ n2 + 19n – 2360 = 0 ⇒ (n + 59)(n – 40) = 0
⇒ n = 40 (∴ n ≠ – 59)
∴ Total no. of instalments = 40
Now, last instalment = 40th instalment
∴ a40 = a + 39d = ₹ 4900

Math Olympiad Test: Arithmetic Progression- 4 - Question 2

### The production of TV in a factory increases uniformly by a fixed number every year. It produced 8000 sets in 6th year and 11300 in 9th year. Find the production in the 6 years.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 2

Since, production of TV in 6th year = 8000
⇒ a6 = 8000 ⇒ a + 5d = 8000 ...(i)
Also, production of TV in 9th year = 11300
⇒ a9 = 11300 or a + 8d = 11300 ...(ii)
Subtracting (i) from (ii), we get
3d = 3300 ⇒ d = 1100
From (i), a = 2500
∴ Production in 6 years, i.e.,
S6 = 6/2[2(2500) + (6 - 1)(1100)] = 31500

Math Olympiad Test: Arithmetic Progression- 4 - Question 3

### Raghav buys a shop for ₹ 120000. He pays half of the amount in cash and agrees to pay the balance in 12 annual instalments of ₹ 5000 each. If the rate of interest is 12% and he pays the interest due on the unpaid amount with the instalment. Find the total cost of the shop.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 3

Amount paid in cash =
Remaining amount = ₹ (120000 – 60000) = ₹ 60000
Amount of 1st instalment

Amount of 2nd instalment

Amount of 3rd instalment

Total amount paid = ₹ 12200 + ₹ 11600 + ₹ 11000 + ... (12 instalments)
which form an A.P. with number of terms,
n = 12, a = 12200 and d = – 600

∴ Total cost of the shop = ₹ 60000 + ₹ 106800 = ₹ 166800

Math Olympiad Test: Arithmetic Progression- 4 - Question 4

A manufacturer of laptop produced 6000 units in 3rd year and 7000 units in the 7th year. Assuming that production increases uniformly by a fixed number every year, find the production in the 5th year.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 4

Since, production of Laptops in 3rd year = 6000
⇒ a3 = 6000 ⇒ a + 2d = 6000 ...(i)
Also production of laptops in 7th year = 7000
a7 = 7000 ⇒ a + 6d = 7000 ...(ii)
Subtracting (i) from (ii), we get
4d = 1000 ⇒ d = 250
From (i), a + 2 (250) = 6000 ⇒ a = 5500
Hence, production in fifth year
a5 = a + 4d = (5500 + 4 (250)) = 6500 units

Math Olympiad Test: Arithmetic Progression- 4 - Question 5

Which of the following statements is INCORRECT?

(a) Sum of n terms of the list of numbers
(b) The common difference of the A.P. given by an = 3n + 2 is 3.
(c) The sum of the A.P. (–5), (–8), (–11), ..., (–230) is –8930.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 5

(a) Given A.P. is

(b) Since, an = 3n + 2
Here, a1 = 3(1) + 2 = 5
a2 = 3(2) + 2 = 8
∴ Common difference = a2 – a1 = 3
(c) Given A.P. is (–5), (–8), (–11), ......, (–230)
⇒ an = a + (n – 1) (–3) ⇒ – 230 = –5 + (n – 1) (–3)
⇒ (-225)/(-3) = (n - 1) ⇒ n = 75 + 1 = 76
⇒ Sn = 76/2((-5) + (230)) = -8930.

Math Olympiad Test: Arithmetic Progression- 4 - Question 6

There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardner will cover in order to water all the trees.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 6

Since, distance of nearest tree from the well = 10 m
Also, each tree is at equal distance of 5 m from the next tree
∴ A.P. formed is 10, 15, 20, ...........
Here, a = 10, d = 5 and n = 25
S25 = (25/2)[2(10) + (25 - 1)5] = 1750
Hence, the total distance the gardener will cover in order to water all the trees = 2 × 1750 = 3500 m.

Math Olympiad Test: Arithmetic Progression- 4 - Question 7

If  is the A.M. between a and b, then find the value of n.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 7

A.M. between a and b =
According to question,

⇒ aan + ban + abn + bbn = 2an + 1 + 2bn + 1
⇒ 2an + 1 + 2bn + 1 – an + 1 – ban – abn – bn + 1 = 0
⇒ an + 1 + bn + 1 – ban – abn = 0
⇒ an(a – b) – bn (a – b) = 0 ⇒ (a – b)(an – bn) = 0
But a – b ≠ 0 ⇒ an – bn = 0 ⇒ an = bn

Math Olympiad Test: Arithmetic Progression- 4 - Question 8

If there are (2n + 1) terms in A.P., then find the ratio of the sum of odd terms and the sum of even terms.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 8

Let a and d be the first term and common difference respectively of the given A.P.
Now, S1 = Sum of odd terms
⇒ S1 = a1 + a3 + a5 + ...+ a2n+1

⇒ S1 = (n + 1) (a + nd)
and, S2 = Sum of even terms
⇒ S2 = a2 + a4 + a6 + ... + a2n ⇒ S2 = n/2[a2 + a2n]
⇒ S2 = n/2 [(a+d)+{a+(2n-1)d}]
⇒ S2 = n(a + nd)
∴ S1 : S2 = (n + 1) (a + nd) : n(a + nd) = (n + 1) : n

Math Olympiad Test: Arithmetic Progression- 4 - Question 9

Fill in the blanks.
(i) If the ratio of sum of n terms of two A.P. is (7n + 1) : (4n + 27), then ratio of their mth terms is P.
(ii) Sum of n odd natural numbers is  Q.
(iii) If sum of n terms of three A.P. are S1, S2, S3. The first term of each is 1 and common differance are 1, 2 and 3 respectively, then

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 9

(i) Let Sn be the sum of n terms of 1st A.P. and S′n be sum of n terms of 2nd A.P.
According to question,

Put n = (2m – 1) in above equation, we get

(ii) A.P. of odd n natural numbers is 1, 3, 5, ............

(iii) We have,

Now,

Math Olympiad Test: Arithmetic Progression- 4 - Question 10

The sum of the third and seventh terms of an A.P. is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Detailed Solution for Math Olympiad Test: Arithmetic Progression- 4 - Question 10

Let a be the first term and d be the common difference of the A.P.
We have, a3 + a7 = 6 and a3a7 = 8
⇒ (a + 2d) + (a + 6d) = 6 and (a + 2d) (a + 6d) = 8
⇒ 2a + 8d = 6 and (a + 2d) (a + 6d) = 8 ⇒ d = ±(1/2)
Case-I: When d =(1/2); a = 1 ⇒ S16 = 76
Case-II: When d = -(1/2); a = 5 ⇒ S16 = 20

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