Math Olympiad Test: Arithmetic Progression- 1 - Class 10 MCQ

Let a be the first term and d be the common difference
a + 2d = 17
a + 6d = 27
Solving these eqns we get a = 12

Let the numbers are a -3d, a - d, a + d, a + 3d
a - 3d + a - d + a + d + a + 3d = 50
⇒ 4a = 50 ⇒ a = 50/4 = 25/2
a + 3d = 4(a - 3d )
⇒ a + 3d = 4a - 12d
⇒ 3a = 15d
⇒ a = 5d
⇒ 25/2 = 5d
⇒ d = 5/2

Let a be the first term and d be the common difference
a + 7d = 31
a + 14d = a + 10d + 16
a + 14d - a - 10d = 16
⇒ d = 4
∴ a = 31 - 7d = 31 - 28 = 3

Let a be the first term and d be common difference
∴ a + d = 8; a + 3d = 14

= 15 [10 + 87] = 15 × 97 = 1455

t_n = 7 - 3n
t₁ = 7 - 3 × 1 = 4 ⇒ a = 4
t₂ = 7 - 3 × 2 = 1
d = 1 - 4 = -3

a = 2 , l = 50

Let the numbers be a - d, a, a + d.
a - d + a + a + d = 12 ⇒ 3a = 12 ⇒ a = 4
and (a - d)³ + a³ + (a + d)³ = 288
a³ - d³ - 3ad(a - d) + a³ + a³ + d³ + 3ad(a + d) = 288
a³ - d³ - 3ad(a - d) + a³ + a³ + a³ + 3ad + 3ad² = 288
⇒ 3a³ + 6ad² = 288 ⇒ 3 × 4³ + 6 × 4 × d² = 288
⇒ 24d² = 288 - 192
⇒ d² = 95/24 = 4
⇒ d = ±2

Given S_n = 3n² + 2n
∴ S₁ = 3 × 1² + 2 × 1 = 5
S₂ = 3 × 2² + 2 × 2 = 16
The sequence is 5, 11, …..
a = 5 , d = 11 - 5 = 6

Given AP is 7, 11, 15, ….139
Here a = 7, d = 11 - 7 = 4, l = 139
l = a + (n - 1) d
⇒ 139 = 7 + (n - 1) 4

⇒ n = 34