Test: Progression (AP And GP)- 5 - CAT MCQ

(a + b)/2 = 25
a + b = 50
√ab = 7
ab = 49
Hence, A can either be 7 or 49.
So, 49 is the answer.

In order to find how many times the alarm rings we need to find the number of numbers below 100
which are not divisible by 2,3, 5 or 7. This can be found by:
100 – (numbers divisible by 2) – (numbers divisible by 3 but not by 2) – (numbers divisible by
5 but not by 2 or 3) – (numbers divisible by 7 but not by 2 or 3 or 5).
Numbers divisible by 2 up to 100 would be represented by the series 2, 4, 6, 8, 10..100 Æ A total
of 50 numbers.
Numbers divisible by 3 but not by 2 up to 100 would be represented by the series 3, 9, 15, 21…
99 Æ A total of 17 numbers. Note use short cut for finding the number of number in this series :
[(last term – first term)/ common difference] + 1 = [(99 – 3)/6] + 1 = 16 + 1 = 17.
Numbers divisible by 5 but not by 2 or 3: Numbers divisible by 5 but not by 2 up to 100 would
be represented by the series 5, 15, 25, 35…95 Æ A total of 10 numbers. But from these numbers,
the numbers 15, 45 and 75 are also divisible by 3. Thus, we are left with 10 – 3 = 7 new numbers
which are divisible by 5 but not by 2 and 3.
Numbers divisible by 7, but not by 2, 3 or 5:
Numbers divisible by 7 but not by 2 upto 100 would be represented by the series 7, 21, 35, 49,
63, 77, 91 Æ A total of 7 numbers. But from these numbers we should not count 21, 35 and 63 as
they are divisible by either 3 or 5. Thus a total of 7 – 3 = 4 numbers are divisible by 7 but not by
2, 3 or 5.

The sum of the interior angles of a polygon are multiples of 180 and are given by (n – 1) × 180
where n is the number of sides of the polygon. Thus, the sum of interior angles of a polygon would
be a member of the series: 180, 360, 540, 720, 900, 1080, 1260
The sum of the series with first term 100 and common difference 10 would keep increasing when
we take more and more terms of the series. In order to see the number of sides of the polygon, we
should get a situation where the sum of the series represented by 100 + 110 + 120… should
become a multiple of 180. The number of sides in the polygon would then be the number of terms
in the series 100, 110, 120 at that point.
If we explore the sums of the series represented by 100 + 110 + 120…
We realize that the sum of the series becomes a multiple of 180 for 8 terms as well as for 9 terms.
It can be seen in: 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 = 1080
Or 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 + 180 = 1260.

The two series till their hundredth terms are 13, 15, 17….211 and 14, 17, 20…311. The common
terms of the series would be given by the series 17, 23, 29….209. The number of terms in this
series of common terms would be 192/6 + 1 = 33. Option (d) is correct.

The maximum score would be the sum of the series 9 + 13 + …. + 389 + 393 + 397 = 98 × 406/2
= 19894. Option (d) is correct.

The side of the first equilateral triangle being 24 units, the first perimeter is 72 units. The second
perimeter would be half of that and so on.
72, 36, 18 …

The 33^rd term of the sequence would be the 17th term of the sequence 3, 9, 15, 21 ….
The 17th term of the sequence would be 3 + 6 × 16 = 99.

Since the four parts of the number are in AP and their sum is 20, the average of the four parts must
be 5. Looking at the options for the largest part, only the value of 8 fits in, as it leads us to think of
the AP 2, 4, 6, 8. In this case, the ratio of the product of the first and fourth (2 × 8) to the product
of the first and second (4 × 6) are equal. The ratio becomes 2:3.

In the course of 2 days the clock would strike 1 four times, 2 four times, 3 four times and so on.
Thus, the total number of times the clock would strike would be:
4 + 8 + 12 + ..48 = 26 × 12 = 312. Option (b) is correct.

The sum of the required series of integers would be given by 7 + 14 + 21 + ….196 = 28 × 101.5 =
2842. Option (b) is correct.

101 + 103 + 105 + … 199 = 150 × 50 = 7500

5 × average of 107 and 253 = 5 × 180 = 900. Option (c) is correct.

If we take the values of a, b, and c as 10,100 and 1000 respectively, we get log a, log b and log c
as 1, 2 and 3 respectively. This clearly shows that the values of log a, log b and log c are in AP.

Trying to plug in values we can see that the infinite sum of the GP 16, 8, 4, 2… is 32 and hence the
third term is 4.

Since the sum of the first five even terms is 15, we have that the 2nd, 4th, 6th, 8th and 10th term of
the AP should add up to 15. We also need to understand that these 5 terms of the AP would also be
in an AP by themselves and hence, the value of the 6th term (being the middle term of the AP)
would be the average of 15 over 5 terms. Thus, the value of the 6th term is 3. Also, since the sum
of the first three terms of the AP is –3, we get that the 2nd term would have a value of –1. Thus, the
AP can be visualized as:
_, -1, _,_,_,3,….
Thus, it is obvious that the AP would be –2, –1, 0, 1, 2, 3. The second term is –1. Thus, option (c)
is correct.