Episode 8 — Aptitude and Reasoning / 8.16 — Arithmetic Progression
8.16.c Arithmetic Progression - Solved Examples
Example 1: Finding the Common Difference
Problem: Find the common difference of the AP: 12, 7, 2, -3, -8, ...
Solution:
d = second term - first term
d = 7 - 12 = -5
Verification:
2 - 7 = -5
-3 - 2 = -5
-8 - (-3) = -5
Common difference d = -5
Example 2: Finding the nth Term
Problem: Find the 20th term of the AP: 4, 9, 14, 19, ...
Solution:
a = 4, d = 9 - 4 = 5, n = 20
a(n) = a + (n - 1)d
a(20) = 4 + (20 - 1)(5)
= 4 + 19 * 5
= 4 + 95
= 99
The 20th term is 99.
Example 3: Which Term is a Given Value?
Problem: Which term of the AP 3, 8, 13, 18, ... is 248?
Solution:
a = 3, d = 5, a(n) = 248
a(n) = a + (n - 1)d
248 = 3 + (n - 1)(5)
245 = (n - 1)(5)
n - 1 = 49
n = 50
248 is the 50th term.
Example 4: Sum of First n Terms
Problem: Find the sum of the first 30 terms of the AP: 5, 11, 17, 23, ...
Solution:
a = 5, d = 6, n = 30
S(n) = n/2 * [2a + (n - 1)d]
S(30) = 30/2 * [2(5) + (30 - 1)(6)]
= 15 * [10 + 174]
= 15 * 184
= 2760
Sum of first 30 terms = 2760.
Example 5: Sum Using First and Last Term
Problem: Find the sum of the AP: 7, 10, 13, ..., 100.
Solution:
a = 7, l = 100, d = 3
Step 1: Find n
n = (l - a)/d + 1 = (100 - 7)/3 + 1 = 93/3 + 1 = 31 + 1 = 32
Step 2: Find sum
S(n) = n/2 * (a + l) = 32/2 * (7 + 100) = 16 * 107 = 1712
Sum = 1712.
Example 6: Finding AP from Two Terms
Problem: The 7th term of an AP is 32 and the 13th term is 62. Find the AP.
Solution:
a(7) = a + 6d = 32 ... (i)
a(13) = a + 12d = 62 ... (ii)
Subtract (i) from (ii):
6d = 30
d = 5
From (i): a + 6(5) = 32 -> a = 32 - 30 = 2
AP: 2, 7, 12, 17, 22, 27, 32, ...
Example 7: Sum of Three Numbers in AP
Problem: The sum of three numbers in AP is 27 and their product is 648. Find the numbers.
Solution:
Let the three numbers be (a - d), a, (a + d)
Sum = (a - d) + a + (a + d) = 3a = 27
a = 9
Product = (a - d)(a)(a + d) = a(a^2 - d^2) = 648
9(81 - d^2) = 648
81 - d^2 = 72
d^2 = 9
d = +/-3
If d = 3: Numbers are 6, 9, 12
If d = -3: Numbers are 12, 9, 6
The numbers are 6, 9, 12.
Example 8: Sum of First n Natural Numbers
Problem: Find the sum of all natural numbers from 1 to 200.
Solution:
Sum = n(n + 1)/2
= 200 * 201 / 2
= 200 * 100.5
= 20100
Example 9: Sum of Even Numbers in a Range
Problem: Find the sum of all even numbers between 50 and 200 (inclusive).
Solution:
Even numbers: 50, 52, 54, ..., 200
a = 50, l = 200, d = 2
n = (200 - 50)/2 + 1 = 150/2 + 1 = 76
S = 76/2 * (50 + 200) = 38 * 250 = 9500
Example 10: Sum of Odd Numbers
Problem: Find the sum of all odd numbers between 1 and 99 (inclusive).
Solution:
Odd numbers: 1, 3, 5, ..., 99
n = (99 - 1)/2 + 1 = 50
Using the formula: Sum of first n odd numbers = n^2
Sum = 50^2 = 2500
Alternatively: S = 50/2 * (1 + 99) = 25 * 100 = 2500
Example 11: Inserting Arithmetic Means
Problem: Insert 5 arithmetic means between 8 and 44.
Solution:
a = 8, b = 44, n = 5
d = (b - a)/(n + 1) = (44 - 8)/(5 + 1) = 36/6 = 6
A(1) = 8 + 6 = 14
A(2) = 8 + 12 = 20
A(3) = 8 + 18 = 26
A(4) = 8 + 24 = 32
A(5) = 8 + 30 = 38
Complete AP: 8, 14, 20, 26, 32, 38, 44
Verification: Sum of AMs = 5 * (8 + 44)/2 = 5 * 26 = 130
14 + 20 + 26 + 32 + 38 = 130 (correct)
Example 12: nth Term from the End
Problem: Find the 8th term from the end of the AP: 3, 7, 11, 15, ..., 199.
Solution:
l = 199, d = 4
nth term from end = l - (n - 1)d
8th from end = 199 - (8 - 1)(4)
= 199 - 28
= 171
Example 13: Finding S(n) from a(n) Expression
Problem: If the nth term of a sequence is a(n) = 4n - 3, find the sum of the first 15 terms.
Solution:
a(n) = 4n - 3
This is an AP with:
a = a(1) = 4(1) - 3 = 1
d = 4 (coefficient of n)
a(15) = 4(15) - 3 = 57
S(15) = 15/2 * (1 + 57) = 15/2 * 58 = 15 * 29 = 435
Example 14: Finding AP from Sum Expression
Problem: If S(n) = 2n^2 + 3n, find the 10th term and the common difference.
Solution:
Method 1 (Using formula):
d = 2A = 2(2) = 4 (A is coefficient of n^2)
a = A + B = 2 + 3 = 5 (B is coefficient of n)
a(10) = 5 + (10-1)(4) = 5 + 36 = 41
Method 2 (Direct calculation):
a(10) = S(10) - S(9)
= [2(100) + 3(10)] - [2(81) + 3(9)]
= [200 + 30] - [162 + 27]
= 230 - 189
= 41
Common difference d = 4
10th term = 41
Example 15: Multiples in a Range
Problem: How many multiples of 9 lie between 200 and 700?
Solution:
First multiple of 9 >= 200:
200/9 = 22.22... -> 23 * 9 = 207
Last multiple of 9 <= 700:
700/9 = 77.77... -> 77 * 9 = 693
n = (693 - 207)/9 + 1 = 486/9 + 1 = 54 + 1 = 55
There are 55 multiples of 9 between 200 and 700.
Example 16: Equidistant Terms Property
Problem: In an AP, the sum of the 4th and 8th terms is 40. Find the sum of the first 11 terms.
Solution:
a(4) + a(8) = 40
By the equidistant property:
a(4) + a(8) = a(1) + a(11) = a(5) + a(7) = a(6) + a(6) = 2*a(6)
So: 2*a(6) = 40 -> a(6) = 20
For 11 terms (odd number), middle term = a(6) = 20
S(11) = 11 * a(6) = 11 * 20 = 220
Example 17: Two APs with Given Sum Ratio
Problem: The ratio of the sums of n terms of two APs is (3n + 1):(4n + 3). Find the ratio of their 7th terms.
Solution:
Let S1(n)/S2(n) = (3n + 1)/(4n + 3)
We know: a(k) = S(2k-1) / (2k-1) for the "single middle term" concept.
For the 7th term: k = 7, so 2k - 1 = 13
The ratio of 7th terms = ratio of S values at n = 13:
a1(7)/a2(7) = (3*13 + 1)/(4*13 + 3) = 40/55 = 8/11
Ratio of 7th terms = 8:11
Example 18: Four Numbers in AP
Problem: Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.
Solution:
Let the four numbers be: (a - 3d), (a - d), (a + d), (a + 3d)
(Common difference = 2d)
Sum = 4a = 28 -> a = 7
Sum of squares = (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 216
Expanding:
= a^2 - 6ad + 9d^2 + a^2 - 2ad + d^2 + a^2 + 2ad + d^2 + a^2 + 6ad + 9d^2
= 4a^2 + 20d^2
4(49) + 20d^2 = 216
196 + 20d^2 = 216
20d^2 = 20
d^2 = 1
d = 1 (taking positive value)
Numbers: (7-3), (7-1), (7+1), (7+3) = 4, 6, 8, 10
Verification: Sum = 28, Sum of squares = 16 + 36 + 64 + 100 = 216 (correct)
Example 19: Salary Problem
Problem: A person gets a starting salary of Rs. 25,000 per month with an annual increment of Rs. 2,000. What is his total earnings in 10 years?
Solution:
Monthly salaries form an AP (changing annually):
Year 1: 25,000/month
Year 2: 27,000/month
...
Year 10: 25,000 + 9(2,000) = 43,000/month
Annual salaries: a = 25,000 * 12 = 3,00,000, d = 2,000 * 12 = 24,000
Total earnings in 10 years:
S(10) = 10/2 * [2(3,00,000) + (10-1)(24,000)]
= 5 * [6,00,000 + 2,16,000]
= 5 * 8,16,000
= Rs. 40,80,000
Example 20: Sum of n Terms Equals Zero
Problem: How many terms of the AP 20, 17, 14, 11, ... must be taken so that the sum is zero?
Solution:
a = 20, d = -3
S(n) = n/2 * [2(20) + (n-1)(-3)] = 0
n/2 * [40 - 3n + 3] = 0
n/2 * [43 - 3n] = 0
Since n != 0:
43 - 3n = 0
n = 43/3 (not an integer)
Wait - let's recalculate:
S(n) = n/2 * [40 - 3(n-1)] = n/2 * [40 - 3n + 3] = n/2 * [43 - 3n]
For S(n) = 0: either n = 0 or 43 - 3n = 0 -> n = 43/3 (not integer)
So the sum never exactly equals zero. Let's check:
S(14) = 14/2 * [43 - 42] = 7 * 1 = 7
S(15) = 15/2 * [43 - 45] = 15/2 * (-2) = -15
The sum is closest to zero at n = 14 (sum = 7) and n = 15 (sum = -15).
Let me re-examine. Actually for the sum to be exactly zero:
n(43 - 3n)/2 = 0 -> n = 43/3 is not an integer.
So the sum never equals exactly zero. The maximum positive sum occurs at n = 14.
Revised Problem Statement: If the question asks "for how many terms is the sum maximum?"
Sum is maximum when terms are still non-negative.
a(n) >= 0 -> 20 + (n-1)(-3) >= 0 -> 23 - 3n >= 0 -> n <= 7.67
So a(7) = 20 - 18 = 2 (positive)
a(8) = 20 - 21 = -1 (negative)
Maximum sum = S(7) = 7/2 * (20 + 2) = 7/2 * 22 = 77
Example 21: Number of Common Terms
Problem: How many terms are common to the APs: 3, 7, 11, 15, ... (up to 100 terms) and 2, 9, 16, 23, ... (up to 50 terms)?
Solution:
AP1: a = 3, d = 4. Terms: 3, 7, 11, 15, ..., a(100) = 3 + 99*4 = 399
AP2: a = 2, d = 7. Terms: 2, 9, 16, 23, ..., a(50) = 2 + 49*7 = 345
Common terms must satisfy both:
3 + 4m = 2 + 7n (for some integers m, n >= 0)
4m - 7n = -1
4m = 7n - 1
The common terms form an AP with d = LCM(4, 7) = 28.
First common term: Check 3, 7, 11, 15, 19, 23...
23 is in AP2 (2, 9, 16, 23). Yes!
So common AP: 23, 51, 79, 107, ..., with d = 28
Last common term <= min(399, 345) = 345
n = (345 - 23)/28 + 1
Let's check: 23 + 28k <= 345
28k <= 322
k <= 11.5
k = 0, 1, 2, ..., 11 -> 12 terms
But let's verify: 23 + 28*11 = 23 + 308 = 331 <= 345 (yes)
23 + 28*12 = 23 + 336 = 359 > 345 (no)
Number of common terms = 12
Example 22: Arithmetic Mean Problem
Problem: The arithmetic mean of two numbers is 34 and their geometric mean is 16. Find the numbers.
Solution:
AM = (a + b)/2 = 34 -> a + b = 68
GM = sqrt(ab) = 16 -> ab = 256
a and b are roots of: x^2 - 68x + 256 = 0
Using quadratic formula:
x = [68 +/- sqrt(68^2 - 4*256)] / 2
= [68 +/- sqrt(4624 - 1024)] / 2
= [68 +/- sqrt(3600)] / 2
= [68 +/- 60] / 2
x = 64 or x = 4
The numbers are 4 and 64.
Example 23: Sum of AP in Terms of Another AP
Problem: If the sum of the first p terms of an AP is q, and the sum of the first q terms is p, find the sum of the first (p + q) terms.
Solution:
S(p) = p/2 * [2a + (p-1)d] = q ... (i)
S(q) = q/2 * [2a + (q-1)d] = p ... (ii)
From (i): p[2a + (p-1)d] = 2q ... (iii)
From (ii): q[2a + (q-1)d] = 2p ... (iv)
Subtract (iv) from (iii):
p[2a + (p-1)d] - q[2a + (q-1)d] = 2q - 2p
2a(p-q) + d[p(p-1) - q(q-1)] = -2(p-q)
2a(p-q) + d[(p^2 - p) - (q^2 - q)] = -2(p-q)
2a(p-q) + d[(p^2 - q^2) - (p - q)] = -2(p-q)
2a(p-q) + d[(p-q)(p+q) - (p-q)] = -2(p-q)
Dividing by (p - q) [assuming p != q]:
2a + d[(p + q) - 1] = -2
2a + (p + q - 1)d = -2 ... (v)
S(p+q) = (p+q)/2 * [2a + (p+q-1)d]
= (p+q)/2 * (-2) [from (v)]
= -(p + q)
Sum of first (p + q) terms = -(p + q).
Example 24: Logs in AP
Problem: If log 2, log(2^x - 1), and log(2^x + 3) are in AP, find x.
Solution:
For three terms in AP: 2 * middle term = sum of first and third
2 * log(2^x - 1) = log 2 + log(2^x + 3)
log(2^x - 1)^2 = log[2(2^x + 3)]
(2^x - 1)^2 = 2(2^x + 3)
Let 2^x = t:
(t - 1)^2 = 2(t + 3)
t^2 - 2t + 1 = 2t + 6
t^2 - 4t - 5 = 0
(t - 5)(t + 1) = 0
t = 5 or t = -1
Since 2^x > 0, t = -1 is rejected.
2^x = 5
x = log(5)/log(2) = log_2(5)
Also check: 2^x - 1 = 4 > 0 and 2^x + 3 = 8 > 0 (valid for log)
x = log_2(5) (approximately 2.32)
Next: 8.16 - Practice MCQs