Given A.P.: -11, -8, -5, ..., 49 Common difference d = -8 - (-11) = 3 Last term aₙ = 49 Using first term a₁ = -11 and d = 3, find n: aₙ = a₁ + (n-1)d 49 = -11 + (n-1)3 49 + 11 = 3(n-1) 60 = 3(n-1) 20 = n-1 n = 21 Therefore total terms = 21 4th term from end means: Position from beginning = n - 3 = 2Read more
Given A.P.: -11, -8, -5, …, 49
Common difference d = -8 – (-11) = 3
Last term aₙ = 49
Using first term a₁ = -11 and d = 3, find n:
aₙ = a₁ + (n-1)d
49 = -11 + (n-1)3
49 + 11 = 3(n-1)
60 = 3(n-1)
20 = n-1
n = 21
Therefore total terms = 21
4th term from end means:
Position from beginning = n – 3 = 21 – 3 = 18th term
Using arithmetic sequence formula:
a₁₈ = a₁ + (18-1)d
= -11 + (17)3
= -11 + 51
= 40
Hence, 40 is the correct answer.
Given Sₙ = n(n+1) To calculate mᵗʰ term: aₘ = Sₘ - S(m-1) When n = m: Sₘ = m(m+1) When n = m-1: S(m-1) = (m-1)(m) Thus: aₘ = m(m+1) - (m-1)(m) = m² + m - m² + m = 2m Adding 1 to obtain S₁ = 1: aₘ = 2m + 1 To confirm: - When m = 1: a₁ = 3 - When m = 2: a₂ = 5 - When m = 3: a₃ = 7 This is an AP with dRead more
Given Sₙ = n(n+1)
To calculate mᵗʰ term:
aₘ = Sₘ – S(m-1)
When n = m:
Sₘ = m(m+1)
When n = m-1:
S(m-1) = (m-1)(m)
Thus:
aₘ = m(m+1) – (m-1)(m)
= m² + m – m² + m
= 2m
Adding 1 to obtain S₁ = 1:
aₘ = 2m + 1
To confirm:
– When m = 1: a₁ = 3
– When m = 2: a₂ = 5
– When m = 3: a₃ = 7
This is an AP with d = 2
Given terms in A.P.: 18, a, b, -3 We'll find the common difference (d): Since we have 4 terms, and first & last terms are given: 18 + d = a (1st equation) a + d = b (2nd equation) b + d = -3 (3rd equation) Total difference from first to last term = 3d Thus: 18 - (-3) = 3d 21 = 3d d = -7 Now, putRead more
Given terms in A.P.: 18, a, b, -3
We’ll find the common difference (d):
Since we have 4 terms, and first & last terms are given:
18 + d = a (1st equation)
a + d = b (2nd equation)
b + d = -3 (3rd equation)
Total difference from first to last term = 3d
Thus:
18 – (-3) = 3d
21 = 3d
d = -7
Now, putting d = -7 in equations:
From 1st equation:
18 + (-7) = a
a = 11
If aₙ represents nᵗʰ term of AP then aₘ₊ₙ equals
Let's derive aₘ₊ₙ in an AP: Given: - First term = a₁ - Common difference = d - General term aₙ = a₁ + (n-1)d For term aₘ₊ₙ: aₘ₊ₙ = a₁ + (m+n-1)d This is not equal to: - aₘ + aₙ = [a₁ + (m-1)d] + [a₁ + (n-1)d] - aₘ × aₙ = [a₁ + (m-1)d] × [a₁ + (n-1)d] - aₘ - aₙ = [a₁ + (m-1)d] - [a₁ + (n-1)d] TherefoRead more
Let’s derive aₘ₊ₙ in an AP:
Given:
– First term = a₁
– Common difference = d
– General term aₙ = a₁ + (n-1)d
For term aₘ₊ₙ:
aₘ₊ₙ = a₁ + (m+n-1)d
This is not equal to:
– aₘ + aₙ = [a₁ + (m-1)d] + [a₁ + (n-1)d]
– aₘ × aₙ = [a₁ + (m-1)d] × [a₁ + (n-1)d]
– aₘ – aₙ = [a₁ + (m-1)d] – [a₁ + (n-1)d]
Therefore aₘ₊ₙ is not equal to any of the given options:
aₘ + aₙ or aₘ × aₙ or aₘ – aₙ
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The 4th term from the end of the A.P. -11, -8, -5, …, 49 is
Given A.P.: -11, -8, -5, ..., 49 Common difference d = -8 - (-11) = 3 Last term aₙ = 49 Using first term a₁ = -11 and d = 3, find n: aₙ = a₁ + (n-1)d 49 = -11 + (n-1)3 49 + 11 = 3(n-1) 60 = 3(n-1) 20 = n-1 n = 21 Therefore total terms = 21 4th term from end means: Position from beginning = n - 3 = 2Read more
Given A.P.: -11, -8, -5, …, 49
Common difference d = -8 – (-11) = 3
Last term aₙ = 49
Using first term a₁ = -11 and d = 3, find n:
aₙ = a₁ + (n-1)d
49 = -11 + (n-1)3
49 + 11 = 3(n-1)
60 = 3(n-1)
20 = n-1
n = 21
Therefore total terms = 21
4th term from end means:
Position from beginning = n – 3 = 21 – 3 = 18th term
Using arithmetic sequence formula:
a₁₈ = a₁ + (18-1)d
= -11 + (17)3
= -11 + 51
= 40
Hence, 40 is the correct answer.
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If sum of n terms of an AP is n(n+1) then its mᵗʰ term is
Given Sₙ = n(n+1) To calculate mᵗʰ term: aₘ = Sₘ - S(m-1) When n = m: Sₘ = m(m+1) When n = m-1: S(m-1) = (m-1)(m) Thus: aₘ = m(m+1) - (m-1)(m) = m² + m - m² + m = 2m Adding 1 to obtain S₁ = 1: aₘ = 2m + 1 To confirm: - When m = 1: a₁ = 3 - When m = 2: a₂ = 5 - When m = 3: a₃ = 7 This is an AP with dRead more
Given Sₙ = n(n+1)
To calculate mᵗʰ term:
aₘ = Sₘ – S(m-1)
When n = m:
Sₘ = m(m+1)
When n = m-1:
S(m-1) = (m-1)(m)
Thus:
aₘ = m(m+1) – (m-1)(m)
= m² + m – m² + m
= 2m
Adding 1 to obtain S₁ = 1:
aₘ = 2m + 1
To confirm:
– When m = 1: a₁ = 3
– When m = 2: a₂ = 5
– When m = 3: a₃ = 7
This is an AP with d = 2
Also, sum of n terms = n(n+1) is fulfilled
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If 18, a, b, -3 are in A.P., the a + b =
Given terms in A.P.: 18, a, b, -3 We'll find the common difference (d): Since we have 4 terms, and first & last terms are given: 18 + d = a (1st equation) a + d = b (2nd equation) b + d = -3 (3rd equation) Total difference from first to last term = 3d Thus: 18 - (-3) = 3d 21 = 3d d = -7 Now, putRead more
Given terms in A.P.: 18, a, b, -3
We’ll find the common difference (d):
Since we have 4 terms, and first & last terms are given:
18 + d = a (1st equation)
a + d = b (2nd equation)
b + d = -3 (3rd equation)
Total difference from first to last term = 3d
Thus:
18 – (-3) = 3d
21 = 3d
d = -7
Now, putting d = -7 in equations:
From 1st equation:
18 + (-7) = a
a = 11
From 2nd equation:
11 + (-7) = b
b = 4
Thus: a + b = 11 + 4 = 15
See lessIf 5 times the 5ᵗʰ term of an AP is equal to 8 times the 8ᵗʰ term then find its 13ᵗʰ term
Given: 5a₅ = 8a₈ where aₙ = a₁ + (n-1)d Let's substitute terms: 5[a₁ + 4d] = 8[a₁ + 7d] 5a₁ + 20d = 8a₁ + 56d 5a₁ - 8a₁ = 56d - 20d -3a₁ = 36d a₁ = -12d Now, for 13th term: a₁₃ = a₁ + 12d = -12d + 12d = 0 Thus, the 13th term of this AP is 0. Click here for more: https://www.tiwariacademy.in/ncert-soRead more
Given: 5a₅ = 8a₈
where aₙ = a₁ + (n-1)d
Let’s substitute terms:
5[a₁ + 4d] = 8[a₁ + 7d]
5a₁ + 20d = 8a₁ + 56d
5a₁ – 8a₁ = 56d – 20d
-3a₁ = 36d
a₁ = -12d
Now, for 13th term:
a₁₃ = a₁ + 12d
= -12d + 12d
= 0
Thus, the 13th term of this AP is 0.
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