AP from Sum Formula
In an AP, if Sₙ = n(4n+1), then find the AP.
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✅ Solution
Step 1: Find first term
Sₙ = n(4n+1)
For n=1: S₁ = 1(4×1+1) = 5
But S₁ = first term = a
∴ a = 5
Step 2: Find second term
For n=2: S₂ = 2(4×2+1) = 2(9) = 18
S₂ = a₁ + a₂ = 5 + a₂ = 18
∴ a₂ = 13
Step 3: Find common difference
d = a₂ – a₁ = 13 – 5 = 8
AP: 5, 13, 21, 29, 37, …
General term: aₙ = 5 + (n-1)8 = 8n – 3
Ratio of mth Terms of Two APs
If the ratio of the sum of first n terms of two A.P.s is (7n + 1) : (4n + 27), find the ratio of their mth terms.
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✅ Solution
Step 1: Use formula for Sₙ
For first AP: Sₙ = n/2 [2a₁ + (n-1)d₁]
For second AP: S’ₙ = n/2 [2a₂ + (n-1)d₂]
Given: Sₙ/S’ₙ = (7n+1)/(4n+27)
Step 2: Cancel n/2
[2a₁ + (n-1)d₁] / [2a₂ + (n-1)d₂] = (7n+1)/(4n+27)
Step 3: Find mth term ratio
mth term of first AP: a₁ + (m-1)d₁
mth term of second AP: a₂ + (m-1)d₂
To get mth terms, put n = 2m-1 in the ratio:
When n = 2m-1, then (n-1) = 2m-2 = 2(m-1)
Ratio of mth terms = [a₁ + (m-1)d₁] / [a₂ + (m-1)d₂]
= [7(2m-1)+1] / [4(2m-1)+27]
= (14m-7+1) / (8m-4+27)
= (14m-6) / (8m+23)
Answer: (14m-6):(8m+23)
Find Number of Terms
Solve: 1 + 4 + 7 + 10 + … + x = 287. Find x and number of terms.
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✅ Solution
Step 1: Identify AP
Given AP: 1, 4, 7, 10, …, x
a = 1, d = 3
Let number of terms = n
Last term x = a + (n-1)d = 1 + (n-1)3 = 3n – 2
Step 2: Use sum formula
Sₙ = n/2 [first term + last term]
287 = n/2 [1 + (3n-2)]
287 = n/2 (3n – 1)
574 = n(3n – 1)
3n² – n – 574 = 0
Step 3: Solve quadratic
3n² – n – 574 = 0
Discriminant = 1 + 4×3×574 = 1 + 6888 = 6889
√6889 = 83
n = [1 ± 83] / 6
n = (1+83)/6 = 84/6 = 14 (positive)
n = (1-83)/6 = -82/6 (reject)
Number of terms = 14
Last term x = 3n – 2 = 3×14 – 2 = 42 – 2 = 40
Answer: n = 14 terms, x = 40
Which Term of AP
Which term of the Arithmetic Progression -7, -12, -17, -22, … will be -82? Is -100 any term of the A.P.?
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✅ Solution
Step 1: Identify AP parameters
AP: -7, -12, -17, -22, …
a = -7, d = -12 – (-7) = -12 + 7 = -5
Step 2: Find term for -82
aₙ = a + (n-1)d = -82
-7 + (n-1)(-5) = -82
-7 -5(n-1) = -82
-5(n-1) = -82 + 7 = -75
n-1 = 15
n = 16
∴ -82 is the 16th term
Step 3: Check if -100 is a term
a + (n-1)d = -100
-7 + (n-1)(-5) = -100
-5(n-1) = -100 + 7 = -93
n-1 = 93/5 = 18.6
n = 19.6
Since n must be a positive integer, -100 is NOT a term of this AP.
Number of Terms for Given Sum
How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.
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✅ Solution
Step 1: Identify AP parameters
AP: 45, 39, 33, …
a = 45, d = 39 – 45 = -6
Let number of terms = n
Sₙ = 180
Step 2: Use sum formula
Sₙ = n/2 [2a + (n-1)d]
180 = n/2 [2×45 + (n-1)(-6)]
180 = n/2 [90 – 6n + 6]
180 = n/2 [96 – 6n]
360 = n(96 – 6n)
360 = 96n – 6n²
6n² – 96n + 360 = 0
Divide by 6: n² – 16n + 60 = 0
Step 3: Solve quadratic
n² – 16n + 60 = 0
(n – 6)(n – 10) = 0
n = 6 or n = 10
Step 4: Explain double answer
Both n=6 and n=10 give sum 180 because:
The AP is decreasing: 45, 39, 33, 27, 21, 15, 9, 3, -3, -9, …
Sum of first 6 terms: 45+39+33+27+21+15 = 180
Sum of first 10 terms: 45+39+33+27+21+15+9+3+(-3)+(-9) = 180
The sum becomes 180 twice because after reaching maximum positive sum, it starts decreasing due to negative terms, and crosses 180 again.
First Negative Term of AP
Which term of the A.P.: 65, 61, 57, 53, … is the first negative term?
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✅ Solution
Step 1: Identify AP parameters
AP: 65, 61, 57, 53, …
a = 65, d = 61 – 65 = -4
Step 2: Find condition for negative term
nth term: aₙ = a + (n-1)d < 0
65 + (n-1)(-4) < 0
65 – 4(n-1) < 0
65 – 4n + 4 < 0
69 – 4n < 0
69 < 4n
n > 69/4 = 17.25
Step 3: Find smallest integer n
Smallest integer n > 17.25 is 18
Check: a₁₈ = 65 + (18-1)(-4) = 65 – 68 = -3
a₁₇ = 65 + (17-1)(-4) = 65 – 64 = 1 (positive)
Answer: 18th term is the first negative term = -3
Triangle Similarity Proof
In the given figure, AD/AE = AC/BD and ∠1 = ∠2. Show that ΔBAE ∼ ΔCAD.
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✅ Solution
Step 1: Given conditions
Given: AD/AE = AC/BD … (i)
Also given: ∠1 = ∠2
Step 2: From Given
∠1 = ∠2 in ΔABD
Therefore, AB = BD [ In Δ Sides Opposite To Equal Angles Are Equal ]
Step 3: Show triangles are similar
From (i): AD/AE = AC/BD [ Given ]
But BD = AB
In ΔCAD and ΔBAE
Therefore, AD/AE = AC/AB. [ BD = AB ]
∠A = ∠A [ Common Angles]
by SAS similarity Criteria
ΔCAD ∼ ΔBAE
Therefore, ΔBAE ∼ ΔCAD
Correct approach:
Given AD/AE = AC/AB (assuming BD should be AB in original)
and ∠A is common to both triangles.
Therefore, by SAS similarity criterion:
ΔBAE ∼ ΔCAD
Isosceles Triangle Proof
If D and E are points on side AB and AC respectively of a ΔABC such that DE || BC and BD = CE. Prove that ΔABC is isosceles.
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✅ Proof
Step 1: Use Basic Proportionality Theorem
Given: DE || BC
By BPT (Thales theorem):
AD/DB = AE/EC … (i)
Step 2: Use given condition BD = CE
Given: BD = CE
Let BD = CE = x
From (i): AD/x = AE/x
∴ AD = AE
Step 3: Prove triangle is isosceles
Now: AB = AD + DB = AD + x
AC = AE + EC = AE + x
Since AD = AE, we get:
AB = AC
Therefore, ΔABC has two equal sides AB and AC.
Hence, ΔABC is isosceles.
Coordinates Division Problem
If the point C(-1, 2) divides internally the line segment joining A(2, 5) and B(x, y) in the ratio 3:4, find the coordinates of B.
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✅ Solution
Step 1: Use section formula
C divides AB in ratio 3:4 (m:n = 3:4)
Coordinates of C = \(\left(\frac{mx₂ + nx₁}{m+n}, \frac{my₂ + ny₁}{m+n}\right)\)
Here: C(-1,2), A(2,5), B(x,y)
m=3, n=4, (x₁,y₁)=(2,5), (x₂,y₂)=(x,y)
Step 2: Apply formula for x-coordinate
-1 = \(\frac{3x + 4×2}{3+4}\)
-1 = \(\frac{3x + 8}{7}\)
-7 = 3x + 8
3x = -15
x = -5
Step 3: Apply formula for y-coordinate
2 = \(\frac{3y + 4×5}{7}\)
2 = \(\frac{3y + 20}{7}\)
14 = 3y + 20
3y = -6
y = -2
Answer: B(-5, -2)
Equidistant Point on x-axis
Write the coordinates of a point P on x-axis which is equidistant from the points A(-2, 0) and B(6, 0).
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✅ Solution
Step 1: Let coordinates of P
Since P lies on x-axis, its y-coordinate = 0
Let P = (x, 0)
Step 2: Use distance formula
Given: PA = PB
PA² = PB²
PA² = (x – (-2))² + (0 – 0)² = (x+2)²
PB² = (x – 6)² + (0 – 0)² = (x-6)²
Step 3: Solve for x
(x+2)² = (x-6)²
x² + 4x + 4 = x² – 12x + 36
4x + 4 = -12x + 36
4x + 12x = 36 – 4
16x = 32
x = 2
Answer: P(2, 0)