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
Given: pᵗʰ term = q. (1) qᵗʰ term = p. (2) Let a₁ be first term and d be common difference. Using (1): a₁ + (p-1)d = q. (3) Using (2): a₁ + (q-1)d = p. (4) Subtracting (4) from (3): (p-q)d = q-p d = -1 Substituting d = -1 in (3): a₁ + (p-1)(-1) = q a₁ - p + 1 = q a₁ = p + q - 1 Now, (p+q)ᵗʰ term = aRead more
Given:
pᵗʰ term = q. (1)
qᵗʰ term = p. (2)
Let a₁ be first term and d be common difference.
Using (1):
a₁ + (p-1)d = q. (3)
Using (2):
a₁ + (q-1)d = p. (4)
Subtracting (4) from (3):
(p-q)d = q-p
d = -1
Substituting d = -1 in (3):
a₁ + (p-1)(-1) = q
a₁ – p + 1 = q
a₁ = p + q – 1
This is the right formula for sum of n terms of an AP. Let's check why: 1) For an AP with first term a₁ and common difference d: Last term (aₙ) = a₁ + (n-1)d 2) In an AP, sum of first and last term = sum of second and second last term =. Hence, Sₙ = n/2(first term + last term) 3) Putting last term:Read more
This is the right formula for sum of n terms of an AP.
Let’s check why:
1) For an AP with first term a₁ and common difference d:
Last term (aₙ) = a₁ + (n-1)d
2) In an AP, sum of first and last term = sum of second and second last term =.
Hence, Sₙ = n/2(first term + last term)
The given sequence is 1/2q, (1-2q)/2q, (1-4q)/2q, ... To find true common difference, let's analyze the pattern: Looking at numerators: 1, (1-2q), (1-4q) Denominator remains constant: 2q In numerator: From 1st to 2nd term: difference is -2q From 2nd to 3rd term: difference is -2q Since denominator iRead more
The given sequence is 1/2q, (1-2q)/2q, (1-4q)/2q, …
To find true common difference, let’s analyze the pattern:
Looking at numerators: 1, (1-2q), (1-4q)
Denominator remains constant: 2q
In numerator:
From 1st to 2nd term: difference is -2q
From 2nd to 3rd term: difference is -2q
Since denominator is 2q,
Common difference = (-2q)/(2q) = 2q
This can be verified:
Starting with first term 1/2q:
– Add 2q: gives (1-2q)/2q (second term)
– Add 2q again: gives (1-4q)/2q (third term)
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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If pᵗʰ term of an AP is q and qᵗʰ term is p then
Given: pᵗʰ term = q. (1) qᵗʰ term = p. (2) Let a₁ be first term and d be common difference. Using (1): a₁ + (p-1)d = q. (3) Using (2): a₁ + (q-1)d = p. (4) Subtracting (4) from (3): (p-q)d = q-p d = -1 Substituting d = -1 in (3): a₁ + (p-1)(-1) = q a₁ - p + 1 = q a₁ = p + q - 1 Now, (p+q)ᵗʰ term = aRead more
Given:
pᵗʰ term = q. (1)
qᵗʰ term = p. (2)
Let a₁ be first term and d be common difference.
Using (1):
a₁ + (p-1)d = q. (3)
Using (2):
a₁ + (q-1)d = p. (4)
Subtracting (4) from (3):
(p-q)d = q-p
d = -1
Substituting d = -1 in (3):
a₁ + (p-1)(-1) = q
a₁ – p + 1 = q
a₁ = p + q – 1
Now, (p+q)ᵗʰ term = a₁ + (p+q-1)(-1)
= (p+q-1) + (p+q-1)(-1)
= p+q-1 – p-q+1
= 0
Therefore, (p+q)ᵗʰ term is 0 is the correct answer.
See lessThe sum of first n terms of an AP is given by
This is the right formula for sum of n terms of an AP. Let's check why: 1) For an AP with first term a₁ and common difference d: Last term (aₙ) = a₁ + (n-1)d 2) In an AP, sum of first and last term = sum of second and second last term =. Hence, Sₙ = n/2(first term + last term) 3) Putting last term:Read more
This is the right formula for sum of n terms of an AP.
Let’s check why:
1) For an AP with first term a₁ and common difference d:
Last term (aₙ) = a₁ + (n-1)d
2) In an AP, sum of first and last term = sum of second and second last term =.
Hence, Sₙ = n/2(first term + last term)
3) Putting last term:
Sₙ = n/2[a₁ + {a₁ + (n-1)d}]
= n/2[2a₁ + (n-1)d]
4) This formula yields:
– When n = 1: S₁ = a₁
– When n = 2: S₂ = 2a₁ + d
– When n = 3: S₃ = 3a₁ + 3d
And so on.
Thus, Sₙ = n/2[2a₁ + (n-1)d] is the right formula.
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The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals
Given first three terms a₁, a₂, a₃ of A.P. are: a₁ = 3y - 1 a₂ = 3y + 5 a₃ = 5y + 1 In an A.P., a₂ - a₁ = a₃ - a₂ So: (3y + 5) - (3y - 1) = (5y + 1) - (3y + 5) Simplifying LHS: 3y + 5 - 3y + 1 = 6 Simplifying RHS: 5y + 1 - 3y - 5 = 2y - 4 As LHS = RHS: 6 = 2y - 4 Adding 4 to both sides: 10 = 2y ThusRead more
Given first three terms a₁, a₂, a₃ of A.P. are:
a₁ = 3y – 1
a₂ = 3y + 5
a₃ = 5y + 1
In an A.P., a₂ – a₁ = a₃ – a₂
So:
(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)
Simplifying LHS:
3y + 5 – 3y + 1 = 6
Simplifying RHS:
5y + 1 – 3y – 5 = 2y – 4
As LHS = RHS:
6 = 2y – 4
Adding 4 to both sides:
10 = 2y
Thus:
See lessy = 2
Thus, 2 is the correct answer.
The common difference of the A.P. is 1/2q, (1-2q)/2q, (1-4q)/2q, … is
The given sequence is 1/2q, (1-2q)/2q, (1-4q)/2q, ... To find true common difference, let's analyze the pattern: Looking at numerators: 1, (1-2q), (1-4q) Denominator remains constant: 2q In numerator: From 1st to 2nd term: difference is -2q From 2nd to 3rd term: difference is -2q Since denominator iRead more
The given sequence is 1/2q, (1-2q)/2q, (1-4q)/2q, …
To find true common difference, let’s analyze the pattern:
Looking at numerators: 1, (1-2q), (1-4q)
Denominator remains constant: 2q
In numerator:
From 1st to 2nd term: difference is -2q
From 2nd to 3rd term: difference is -2q
Since denominator is 2q,
Common difference = (-2q)/(2q) = 2q
This can be verified:
Starting with first term 1/2q:
– Add 2q: gives (1-2q)/2q (second term)
– Add 2q again: gives (1-4q)/2q (third term)
Hence, 2q is the correct answer.
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If the roots of the quadratic equation x² – (a + 1)x + a = 0 are equal, then the value of a is:
Given quadratic equation: x² - (a + 1)x + a = 0 For equal roots, discriminant should be zero: b² - 4ac = 0 Here: a = 1 (coefficient of x²) b = -(a + 1) c = a Substituting in discriminant: [-(a + 1)]² - 4(1)(a) = 0 (a + 1)² - 4a = 0 a² + 2a + 1 - 4a = 0 a² - 2a + 1 = 0 (a - 1)² = 0 Therefore: a = 1 TRead more
Given quadratic equation: x² – (a + 1)x + a = 0
For equal roots, discriminant should be zero:
b² – 4ac = 0
Here:
a = 1 (coefficient of x²)
b = -(a + 1)
c = a
Substituting in discriminant:
[-(a + 1)]² – 4(1)(a) = 0
(a + 1)² – 4a = 0
a² + 2a + 1 – 4a = 0
a² – 2a + 1 = 0
(a – 1)² = 0
Therefore:
a = 1
This can be verified by substituting a = 1 in original equation:
x² – 2x + 1 = 0
(x – 1)² = 0
x = 1 (repeated root)
Hence, 1 is the correct answer.
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