Arithmetic Progression: Solved Examples

Sol : Option B
Explanation The smallest three digit number that will leave a remainder of two when divided by three is 100.
The next no. that will leave a remainder of two when divided by three is 103, 106,
The largest three digit number that will leave a remainder of two when divided by three is 997.
So, it is an Arithmetic Progression with the first term being 100 and the last term being 997 and common difference being three.
We know that in an Arithmetic Progression, the nth term an = a1 + (n - 1) d
In this particular case, therefore, 997 = 100 + (n - 1)* 3
i.e., 897 = (n - 1) * 3
Therefore, n - 1 = 299
Or n = 300.
So, the sum of the Arithmetic Progression will be = n/2[2a+ (n−1) d] =164,850

Sol : Option D
Explanation: The positive integers, which are exactly divisible by five, are 5, 10, 15, ..., 1000
Out of these 10, 20, 30,,.... 1000 are divisible by two.
Therefore, we will have to find the sum of all the positive integers 5, 15, 25, ...., 995
If n is the no. of terms in it, then sequence is
995 = 5 + 10(n - 1) ⇒ 990 = 10n – 10
1000 = 10n
Thus, n = 100.
Thus the sum of arithmetic progression series = (n/2) (a + l) = (100/2) (5 + 995) = 50000.

Sol : Option D
Explanation: Let the numbers be a - d, a, a + d
Then a - d + a + a + d = 21
3a = 21
a = 7
and (a - d)(a + d) = 45
a² – d² = 45
d² = 4
d = +2
Hence, the numbers are 5, 7 & 9 when d = 2 and 9, 7 & 5 when d = -2. In both the given cases numbers are the same.

Sol : Option B
Explanation: Let the cost of the equipment be Rs. 100.
Now, percentages of depreciation at the end of 1st, 2nd, 3rd years are 15, 13.5, 12, respectively which are in A.P., with a = 15 and d = - 1.5.
Hence, percentage of depreciation in the tenth year = a + (10-1) d = 15 + 9 (-1.5) = 1.5
Also total value depreciated in the 10 years = 15 + 13.5 + 12 + ... + 1.5 = 82.5
Hence, value of equipment at end of 10 years=100 - 82.5 = 17.5.
The total cost being Rs. 6, 00,000/100 * 17.5 = Rs. 1, 05,000.

Sol : Option A
Explanation: The third term t₃ = a + 2d
The ninth term t₉ = a + 8d
t₃ + t₉ = 2a + 10d = 8
Sum of 1st 11 terms of an AP is given by
S₁₁ = 11/2 [2a +10d]
S₁₁ = 11/2 * 8 =44

Sol : Option B
Explanation: 12 = a + 2d
106 = a + 49d
So, 106-12 = 47d
Or, 94 = 47d
So, d = 2
Hence, a = 8
And, n₂₉ = 8 + 28*2 = 64.

Sol : Option C
Explanation: Number of two-digit positive integers divisible by 4
The smallest two-digit positive integer divisible by 4 is 12. The largest two-digit positive integer divisible by 4 is 96.
All the two-digit positive integers are terms of an Arithmetic progression with 12 being the first term and 96 being the last term.
The common difference is 4.
The nth term an = a1 + (n - 1) d, where a1 is the first term, 'n' number of terms and’d’ is the common difference.
So, 96 = 12 + (n - 1) * 4
Or 84 = (n - 1) * 4
Or (n - 1) = 21
Hence, n = 22.
i.e., there are 22 two-digit positive integers that are divisible by 4.

Number of two-digit positive integers divisible by 9
The smallest two-digit positive integer divisible by 9 is 18. The largest two-digit positive integer divisible by 9 is 99.
All the two-digit positive integers are terms of an Arithmetic progression with 18 being the first term and 99 being the last term.
The common difference is 9.
The nth term an = a1 + (n - 1) d, where a1 is the first term, 'n' number of terms and 'd' the common difference.
So, 99 = 18 + (n - 1) * 9
Or 81 = (n - 1) * 9
Or (n - 1) = 9
Hence, n = 10.
i.e., there are 10 two-digit positive integers that are divisible by 9.
Removing double count of numbers divisible by 4 & 9
Numbers such as 36 and 72 are multiples of both 4 & 9 and have therefore been counted in both the groups.
There are 2 such numbers.
Hence, number of two-digit positive integers divisible by 4 or 9
= Number of two-digit positive integers divisible by 4 + Number of two-digit positive integers divisible by 9- Number of two-digit positive integers divisible by 4 and 9
= 22 + 10 - 2 = 30

Sol : Option A
Explanation: Of all the 4 terms in AP, the 2^nd term is 17, which is an odd number.
The common difference has to be either even or odd.
Possibility 1: If the common difference is odd, 1st term will be even, 3rd term will be even and the 4th term will be odd.
i.e., two of the terms of the sequence are odd & two are even.
Sum of two odd numbers and two even numbers is even.
Possibility 2: If common difference is even, all 4 terms will be odd.
Sum of 4 odd numbers is even.
So, irrespective of whether common difference is odd or even, the sum of four terms is even.
Hence, the sum will be divisible by 2

Sol : Option B
Explanation: The sum of any set of numbers = Average of the numbers * number of terms
1. Step 1: Compute Average
Average of terms of an arithmetic sequence = [first term + last term/2]
Average of this sequence = [−64 −100]/ 2 = -82
2. Step 2: Compute Sum:
Sum = Average * number of terms
We have computed the no. of terms in the text book approach. Number of terms = 19.
Therefore, sum = (-82) * 19 = -1558

Sol : Option B
Explanation: Because log 2, log (2x -1) and log (2x + 3) are in an AP
2 log (2x - 1) = log 2 + log (2x + 3)
Using the power rule of log, we can write 2 log (2x - 1) = log (2x - 1)²
Therefore, we can write the equation as log (2x - 1)² = log 2 + log (2x + 3)
Using product rule, we can write the right hand side of the equation as follows:
log 2 + log (2x + 3) = log 2 (2x + 3)
The modified equation after applying the power rule and the product rule of log is log (2x - 1)² = log 2 (2x + 3)
Removing log on both sides of the equation, we get
(2x - 1)² = 2 (2x + 3)
Or 4x² - 2×2x + 1 = 2 × 2x + 6
Or 4x² - 4 × 2x - 5 = 0
Let y = 2x
So, the equation can be written as y² - 4y - 5 = 0
Factorizing, we get (y - 5) (y + 1) = 0.
So, y = 5 or y = -1
So, 2x = 5 or 2x = -1
2x cannot be negative. S, 2x = 5
If 2x = 5, expressed in terms of a log, it becomes x = log25