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

1^st instalment = ₹ 1000
2^nd instalment = ₹ 1000 + ₹ 100 = ₹ 1100
3^rd 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 = 40^th instalment
∴ a₄₀ = a + 39d = ₹ 4900

Since, production of TV in 6^th year = 8000
⇒ a₆ = 8000 ⇒ a + 5d = 8000 ...(i)
Also, production of TV in 9^th 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

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

Amount of 2^nd instalment

Amount of 3^rd 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

Since, production of Laptops in 3^rd year = 6000
⇒ a₃ = 6000 ⇒ a + 2d = 6000 ...(i)
Also production of laptops in 7^th 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

(a) Given A.P. is 

(b) Since, a_n = 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)
⇒ a_n = a + (n – 1) (–3) ⇒ – 230 = –5 + (n – 1) (–3)
⇒ (-225)/(-3) = (n - 1) ⇒ n = 75 + 1 = 76
⇒ S_n = 76/2((-5) + (230)) = -8930.

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.

A.M. between a and b = 
According to question,
 
⇒ aaⁿ + baⁿ + abⁿ + bbⁿ = 2a^{n + 1} + 2b^{n + 1}
⇒ 2a^{n + 1} + 2b^{n + 1} – a^{n + 1} – baⁿ – abⁿ – b^{n + 1} = 0
⇒ a^{n + 1} + b^{n + 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

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 + ...+ a_2n+1

⇒ S₁ = (n + 1) (a + nd)
and, S₂ = Sum of even terms
⇒ S₂ = a₂ + a₄ + a₆ + ... + a_2n ⇒ S₂ = n/2[a₂ + a_2n]
⇒ 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

(i) Let S_n be the sum of n terms of 1^st A.P. and S′_n be sum of n terms of 2^nd 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, 

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