Math Olympiad Test: Arithmetic Progression- 2 - Class 10 MCQ

Given AP is 11 + 13 + 15 + …..+ 99
Here a = 11, d = 13 - 11 = 2, l = 99
l = a + (n - 1)d
⇒ 99 = 11 + (n - 1)2

⇒ n - 1 = 44
⇒ n = 45

= 45 × 55 = 2475

loga, logab, logab², logab³, ….
loga, loga + logb, loga + 2logb, loga + 3logb
t₁ = loga, d = loga + logb - loga = logb
t_n = t₁ + (n - 1)d = loga + (n - 1)logb
= loga + logb^n-1
= logab^n-1

S₂ - S₁ = 2P + Q - P = P + Q
P, P + Q ….
Common difference = Q

Let the numbers be a - d, a, a + d
Now a - d + a + a +d = 27 ⇒ a = 9
and (a - d) a (a + d) = 405
a(a² - d²) = 405
9(9² - d²) = 405
⇒ 81 - a² = 45
⇒ d² = 36 ⇒ d = ±6
The numbers are 3, 9, 15.

The three-digit natural numbers which are multiples of 7 are
105, 112, 119,…, 994
a₂ – a₁ = 112 – 105 = 7
a₃ – a₂ = 112 – 105 = 7
∵ a₃ – a₂ = a₂ – a₁ = 7
Therefore, the series is in AP
Here, a = 105, d = 7 and a_n = 994
We know that,
a_n = a + (n – 1)d
⇒ 994 = 105 + (n – 1)7
⇒ 994 – 105 = (n – 1)7
⇒ 889 = (n – 1)7
⇒ 127 = (n – 1)
⇒ n = 128
Now, we have to find the sum of this AP

⇒ S₁₂₈ = 64[210 + 127 × 7]
⇒ S₁₂₈ = 64[1099]
⇒ S₁₂₈ = 70336
Hence, the sum of all three-digit numbers which are multiples of 7 is 70336.

Given: Sum of first 7 terms, S₇ = 10
and Sum of the next 7 terms = 17
⇒ Sum of 8^th to 14^th terms = 17
⇒ Sum of first 14 terms – Sum of first 7 terms = 17
⇒ S₁₄ – S₇ = 17
⇒ S₁₄ – 10 = 17
⇒ S₁₄ = 27

Let a be the first term and d be the common difference.

2am + m(m - 1)d = 2n (1)

2an + n(n - 1)d = 2m (2)
Subtracting (2) from eqn (1), we get
2a(m - n) + [(m² - n²) - (m - n)]d = 2(n - m)
⇒ (m - n)[2a + (m + n - 1) d] = 2(n - m)
2a + (m + n - 1)d = -2

Let x be the first term and d be the common difference.
p^th term = a
x + (p - 1)d = a ...(1)
q^th term = b ⇒ x + (q - 1)d = b ...(2)
r^th term = c ⇒ x + (r - 1)d = c ...(3)
∴ a(q - r) + b(r - p) + c(p - q)
= x[q - r + r - p + p - q] + d[(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (q - q)]
⇒ a(q - r) + b(r - p) + c(p - q) = 0

t_k = 5k + 1
t₁ = 5 × 1 + 1 = 6
t_n = 5n + 1

3, 7, 11, 15…..
a = 3, d = 4, Sn = 406


⇒ n[6 + 4n - 4] = 812
⇒ 4n² + 2n = 812
⇒ 2n² + n = 406
⇒ 2n² + 29n - 28n - 406 = 0
⇒ n(2n + 29) × 14(2n + 29) = 0
⇒ (n - 14) (2n + 29) = 0

Here 4k - 1 - (5k + 2) = (k + 2) - (4k - 1)
⇒ 4k - 1 - 5k - 2 = k + 2 - 4k + 1
⇒ -k - 3 = -3k + 3
⇒ 2k = 6 ⇒ k = 3

Let a be the first term and d be the common difference


Subtracting eqn (2) from eqn (1)

a = 4, d = 7 - 4 = 3, tn = 31
∴ tn = a + (n - 1)d
⇒ 31 = 4 + (n - 1) 3
⇒
⇒ n - 1 = 9 ⇒ n = 10