Progressions (A.P., G.P., H.P.) Questions and Answers

Answer: B

The 100th common term is 597.

The first series is the odd numbers from 3, and the second is the multiples of 3. Their common terms are the odd multiples of 3: 3, 9, 15, 21, ..., an arithmetic progression with first term 3 and common difference 6.

\[ \begin{aligned} T_n &= a + (n-1)d \\ T_{100} &= 3 + 99 \times 6 \\ &= 597 \end{aligned} \]

Using \(n\) instead of \(n-1\) gives \(3 + 600 = 603\). 595 and 607 are not odd multiples of 3, and 605 is not a multiple of 3, so none of them can be a common term.

Answer: C

There are 15 scooters in the second half.

Each car is followed by a growing number of scooters: car 1 then 1 scooter, car 2 then 2 scooters, and so on. After the 7th car and its 7 scooters, the row holds 7 + (1 + 2 + ... + 7) = 7 + 28 = 35 vehicles, so the 36th vehicle is the 8th car.

Numbering the positions: cars stand at 1, 3, 6, 10, 15, 21, 28 and 36, and every other position holds a scooter. The second half is positions 19 to 36. Its scooters are at 19 and 20 (2, after car 5), 22 to 27 (6, after car 6) and 29 to 35 (7, after car 7): 2 + 6 + 7 = 15.

As a check, the row has 28 scooters in all, so the first half holds 13, and 13 + 15 = 28.

Answer: A

The flower bed has 10 rows.

The rows form an arithmetic progression with first term 23 and common difference -2, ending at 5. The number of terms is:

\[ \begin{aligned} n &= \frac{\text{first} - \text{last}}{\text{difference}} + 1 \\ &= \frac{23 - 5}{2} + 1 = 9 + 1 = 10 \end{aligned} \]

Check: 23, 21, 19, 17, 15, 13, 11, 9, 7, 5 is ten rows. Forgetting the +1 gives 9; 15, 17 and 18 do not fit the sequence.

Answer: A

The number in the list is 179.

The list 5, 11, 17, 23, ... increases by 6, so every term is one less than a multiple of 6: the nth term is 6n - 1. A number belongs to the list if adding 1 gives a multiple of 6.

\[ \begin{aligned} 179 + 1 &= 180 = 6 \times 30 \\ 189 + 1 &= 190,\ 169 + 1 = 170,\ 159 + 1 = 160 \end{aligned} \]

Only 180 is divisible by 6, so 179 is the 30th term. 190, 170 and 160 leave remainders when divided by 6, so 189, 169 and 159 are not in the list.

Answer: B

There are 128 three-digit numbers divisible by 7.

The smallest three-digit multiple of 7 is 105 (7 × 15) and the largest is 994 (7 × 142). Counting the multiples from the 15th to the 142nd:

\[ \begin{aligned} n &= \frac{994 - 105}{7} + 1 \\ &= 127 + 1 = 128 \end{aligned} \]

Equivalently, 142 - 15 + 1 = 128. Forgetting to add 1 gives 127, the most common slip; 129 and 130 overcount.

Answer: C

The sum is 400.

The sum of the first n odd natural numbers is always \(n^{2}\): 1 = 1, 1 + 3 = 4, 1 + 3 + 5 = 9, and so on. For n = 20:

\[ 1 + 3 + 5 + \dots + 39 = 20^{2} = 400 \]

The same result comes from the arithmetic series formula: 20 terms with average (1 + 39)/2 = 20 give 20 × 20 = 400. 210 is the sum of the first 20 natural numbers, and 420 is the sum of the first 20 even numbers, 2 + 4 + ... + 40.

Answer: D

The eighth term is 31.

The \(n\)th term is the difference of consecutive sums:

\[ \begin{aligned} T_{8} &= S_{8} - S_{7} \\ &= (2 \cdot 64 + 8) - (2 \cdot 49 + 7) \\ &= 136 - 105 = 31 \end{aligned} \]

Check: \(S_{1} = 3\) and \(S_{2} = 10\), so the progression is 3, 7, 11, ... with common difference 4, and \(T_{8} = 3 + 7 \times 4 = 31\). 136 is \(S_{8}\), the sum of eight terms, not the eighth term.

Answer: B

The value of \(x\) is 4.

If 3, \(x\), 6 are in harmonic progression, their reciprocals are in arithmetic progression, so \(\frac{1}{x}\) is the average of \(\frac{1}{3}\) and \(\frac{1}{6}\):

\[ \begin{aligned} \frac{2}{x} &= \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \\ x &= 4 \end{aligned} \]

Equivalently, the harmonic mean is \(\frac{2ab}{a+b} = \frac{36}{9} = 4\). 4½ is the arithmetic mean of 3 and 6, the trap; 5 and 5½ fit neither mean.

Answer: D

She received 2046 apples.

The apples form a geometric progression 2, 4, 8, ... with first term 2 and common ratio 2, for 10 terms:

\[ \begin{aligned} S_{10} &= \frac{2(2^{10} - 1)}{2 - 1} \\ &= 2(1024 - 1) = 2046 \end{aligned} \]

1024 is the number given on the tenth day alone (\(2^{10}\)), not the total; 2060 and 1760 do not come from the series.

Answer: D

The sum t20 to t30 has 11 terms and equals 0, so the middle term t25 is 0, which makes (A) true. Terms t10, t20, t30 and t40 are equally spaced, so they form an A.P. (B), and for n1 + n2 = 50 the terms are placed symmetrically about t25, so their sum is 0 (C).

Since t25 = a + 24d = 0, a = -24d, so if d is positive then a is negative. Hence (D) is false.

If the sum of 11 consecutive terms of an arithmetic progression is zero, the middle term is zero, so t25 = a + 24d = 0 and a = -24d. For positive d this makes a negative, not positive, which is why statement D is false. Statement C follows from the symmetry about t25.