2IIM Quant Notes  Arithmetic and Geometric Progression

rajesh_balasubramanian
Director, 2IIM Online CAT Preparation  IIT Madras  IIM Bangalore  CAT 100th percentile  CAT 2011, 2012 and 2014.
With some simple but very powerful ideas, one can cut down on a lot of working when it comes to arithmetic progression(AP) & geometric progression(GP). For example, anchoring a progression around its middle term can be very useful. Reinforce these ideas with the following questions.
Second term of a geometric progression is 1000 and the common ratio is where n is a natural number. P_{n} is the product of n terms of this GP. P_{6} > P_{5} and P_{6} > P_{7}, what is the sum of all possible values of n?
 4
 9
 5
 13
Common ratio is positive, and one of the terms is positive = > All terms are positive
P_{6} = P_{5} * t_{6} = > If P_{6} > P_{5}, t_{6} > 1
P_{7} = P_{6} * t_{7} = > If P_{6} > P_{7}, t_{7} < 1
t_{6} = t_{2} * r^{4} = 1000r^{4};
t_{7} = t_{2} * r^{5} = 1000r^{5}
1000r^{4} > 1 and 1000r^{5} < 1
1/r^{4 } < 1000 and 1/r^{5 } > 1000.
1/r = n.
n^{4} < 1000 and n^{5} > 1000, where n is a natural number
n^{4} < 1000 = > n < 6
n^{5} > 1000 = > n ≥ 4
n could be 4 or 5. Sum of possible values = 9 Correct Answer: 9Sum of first 12 terms of a geometric progression is equal to the sum of the first 14 terms in the same GP. Sum of the first 17 terms is 92, what is the third term in the geometric progression?
 92
 92
 46
 231
Sum of first 12 terms is equal to sum of first 14 terms.
Sum of first 14 terms = Sum of first 12 terms + 13^{th} term + 14^{th} term
= > 13^{th} term + 14^{th} term = 0
Let us assume 13^{th} term = k, common ratio = r.
14^{th} term will be kr.
k + kr = 0
k (1 + r) = 0
= > r = 1 as k cannot be zero
Common ratio = 1.
Now, if the first term of this geometric progression is a, second term would be a, third would be a and so on
The geometric progression would be a, a, a, a, a, a,...
Sum to even number of terms = 0
Sum to odd number of terms = a
Sum to 17 terms is 92 = > a = 92
Third term = a = 92
Correct Answer: 92Sum of first 25 terms in arithmetic progression is 525, sum of the next 25 terms is 725, what is the common difference
 8/25
 4/25
 6/25
 1/25
The sum of the first 25 terms, S_{25} = 525
The sum of the next 25 terms, K_{25} = sum of next 25 terms = 725
a_{26} = a_{1} + 25d
a_{27} = a_{2} + 25d
K_{25} = a_{26} + a_{27} + ...... + a_{50}
Or K_{25} = a_{1} + 25d + a_{2} + 25d + ...... + a_{25} + 25d
Or K_{25} = a_{1} + a_{2} + ...... + a_{25} + 25(25d)
Or K_{25} = S_{25} + 25(25d)
i.e., 725 = 525 + 25 x 25d
200 = 25 x 25d
8 = 25d
d = 8/25Let the n^{th} term of arithmetic progression be defined as t_{n}, and sum up to 'n' terms be defined as S_{n}. If t_{8} = t_{16} and t_{3} is not equal to t_{7}, what is S_{23}?
 23(t_{16}  t_{8})
 0
 23t_{11}
 Cannot be determined
t_{8}=t_{16}. This can happen under two scenarios t_{8} = t_{16} or t_{8} = – t_{16}.
If t_{8} = t_{16}, the common difference would be 0 suggesting that t_{3} would be equal to t_{7}. However, we know t_{3} is not equal to t_{7}, so the common difference cannot be zero.
This tells us that t_{8} = – t_{16} Or, t_{8} + t_{16} = 0.
If t_{8} + t_{16} = 0, then t_{12} = 0. t_{12} = t_{8} + 4d, and t_{16} – 4d So, t_{12} = ( t_{8 + }t_{16} ) / 2
For any two terms in an AP, the mean is the term right in between them. So, t_{12} is the arithmetic mean of t_{8} and t_{16}.
So, t_{12} = 0.
Now, S_{23} = 23 × t_{12}. We know that average of n terms in an A.P. is the middle term. This implies that sum of n terms in an A.P., is n times the middle term. So, S_{23} = 0.a, b, c, d and e are 5 distinct numbers that from an arithmetic progression. They are not necessarily consecutive terms but form the first 5 terms of the arithmetic progression. It is known that c is the arithmetic mean of a and b, and d is the arithmetic mean of b anc c. Which of the following statements are true?
i. Average of all 5 terms put together is c.
ii. Average of d and e is not greater than average of a and b.
iii. Average of b and c is greater than average of a and d. i and ii only
 ii and iii only
 all three statements are true
 i and iii only
I. c is the arithmetic mean of a and b = > c lies in between a and b. And it lies exactly in between the two terms. As in the number of terms between a and c should be equal to number of terms between b and c.
a, c, b could be the
1st 2nd and 3rd terms respectively or
1st, 3rd and 5th respectively, or
2nd, 3rd, 4th respectively, or
3rd, 4th, 5th respectively.
The terms could also be the other way around. As in, b, c, a could be the 1st 2nd and 3rd terms respectively, or the 1st, 3rd and 5th respectively, and so on. This is a very simple but very powerful idea.II. Now, d is the arithmetic mean of b and c. = > d lies between b and c. Using statements I and II we can say that a, c, b have to be 1st, 3rd and 5th or 5th, 3rd and 1st as there is an element between b and c also.
So, c is the third term. a and b are 1st and 5th in some order.
b ___ c ___ a or a ___ c ___ b
d is the arithmetic mean of b and c.
Possible arrangements are:
b d c e a or a e c d bStatement (I): The average of all 5 terms put together is c. c is the middle term. So this is true.
Statement (II): the average of d and e is not greater than average of a and b. Average of a, b is c. d and e are the 2nd and 4th terms of this sequence (in some order). So, their average should also be equal to c. So, both these are equal. So, this statement is also true.
Statement (III): The average of b and c is greater than average of a and d. The average of b and c is d. The average of a and d could be greater than or less than d. So, this need not be true.
Correct Answer: i and ii are trueConsider a, b, c in a G.P. such that a + b + c = 15. The median of these three terms is a, and b = 10. If a > c, what is the product of the first 4 terms of this G.P.?
 40000
 32000
 8000
 48000
Median is the first term = > Common ratio has to be negative. why?
Let us see why this is true.
When a > 0,
If r > 1, this will be an increasing G.P.
If r lies between (0, 1), this will be a decreasing G.P.
In both cases, the middle term will be the median. If a < 0, the order will be the other way around, but the middle term would still be the median.
If the middle term is not the median, we can say that r < 0. Now, let us go the solution
b = 10, a and c should be negative. Solution, a + b + c cannot be 15.
a + b + c = –15
10/r + 10 + 10r = 5 = > 2/r + 2r = 5
Solving the quadratic, we will get r = 1/2 or 2
The sequence is either – 5, 10, – 20 or – 20, 10, – 5.
a > c = > the sequence has to be – 5, 10, – 20.
The product of the first 4 terms = – 5 × 10 × –20 × 40 = 40000.
Correct Answer: 40000If 4 times the 4th term of an A.P. is equal to 9 times the 9th term of the A.P., what is 13 times the 13th term of this A.P.?
 7 times the 13th term
 0
 13 times the 7th term
 4 times the 4th term + 9 times the 9th term
4t _{4} = 9t _{9} , we need to find t_{13}
4(a + 3d) = 9(a + 8d)
4a + 12d = 9a + 72d
= > 5a + 60d = 0
= > a + 12d = 0
= > t_{13} = 0
= > 13 × t"_{13}" = 0
As a simple rule, if n × t_{n} = m × t_{m}, then t_{m+n} = 0. See, if you can prove this.
Answer choice (b).
Correct Answer: 0Sequence P is defined by p_{n} = p_{n1} + 3, p_{1} = 11, Sequence Q is defined as q_{n} = q_{n1} – 4, q_{3} = 103. If p_{k} > q_{k+2}, what is the smallest value k can take?
 6
 11
 14
 15
Sequence P is an A.P. with a = 11, and common difference 3.
So, P_{k} = 11 + (k – 1)3.
Sequence Q is an A.P. with third term 103 and common difference – 4.
t_{3} = a + 2d
103 = a + 2 (– 4) or a = 111
q_{k+2} = 111 + (k +1) (– 4)
q_{k + 2} = 111 – 4k – 4 = 107 – 4k
p_{k} > q_{k + 2}
11 + (k–1)3 > 107 – 4k
8 + 3k > 107 – 4k
7k > 99
k > 99/7
k has to be an integer, so smallest value k can take is 15.
Answer choice (d).
Correct Answer: 15The sum of 2n terms of A.P. {1, 5, 9, 13…..} is greater than sum of n terms of A.P. = {56, 58, 60..…}. What is the smallest value n can take?
 9
 10
 12
 14
First A.P., a = 1, d = 4
S_{2n} = 2n/2 [2 x 1+(2n −1)4]
S_{2n} = n(2 + 8n – 4) = n(8n – 2) = 8n^{2} – 2n
For the second arithmetic progression, a = 56, d = 2
S_{n} = n/2 [2 x 56+(n −1)2]
S_{n} = n/2 [112+2n −2] = n/2 (110+2n)
The sum of 2n terms of AP {1, 5, 9, 13, …..} is greater than sum of n terms of A.P. = {56, 58, 60, …}
8n^{2} – 2n > n/2 (110+2n)
16n^{2} – 4n > 110n + 2n2
14n^{2} > 114n
7n > 57
n > 57/7
The smallest value n can take = 9. Answer choice (a).a, b, c and d are in A.P., What can we say about terms bcd, acd, abd and abc?
 They are also in A.P.
 They are also in H.P.
 They are also in G.P.
 They are not in an A.P., G.P. or H.P.
a, b, c and d are in A.P.
Dividing through out this A.P. by abcd, we get
a/abcd , b/abcd , c/abcd , d/abcd are in A.P.
1/bcd , 1/acd , 1/abd , 1/abc are in A.P.
= > bcd, acd, abd and abc are in H.P.
Answer choice (b).Second term in an arithmetic progression is 8 and the 8th term is 2 more than thrice the second term. Find the sum up to 8 terms of this AP.
 124
 108
 96
 110
Given, Second term of an AP is 8 = > a+d =8, 8th term is 2 more than thrice the second term = > a+7d = 2 +3(a+d) = 2+3*8 =26 .
a+d =8  1
a+7d = 26  2
Solving for d in the two equations, we get d = 3 and a =5. Sum of n terms in an AP = n/2 x [2a + (n1)d]
= > Sum upto 8 terms in this arithmetic progression = 8/2 * [2*5 + (81)3] = > 4*[10+21] = 4*31 = 124.If S_{n }= n^{3 }+ n^{2}+ n^{}+ 1 where S_{n }denotes the sum of the first n terms of a series and t_{m }= 291, then m is equal to?
 24
 30
 26
 20
The pattern is as follows : 2, 23, 234...
S_{n}S_{n1} = t_{n}
Substitute m instead of n
S_{m}S_{m1} = t_{m}
We know that S_{n }= n^{3 }+ n^{2}+ n^{}+ 1
Hence m^{3}+m^{2}+m+1
m^{3}+m^{2}+m+1[(m1)^{3}+(m1)^{2}+(m1)+1]=291
m^{3}+m^{2}+m+1[(m)^{3}(3m)^{2}+3m1+(m)^{2}2m+1+m1+1]=291
1+3m^{2}+3m+12m1=291
3m^{2}+m290=0
3m^{2}m+290=0
Solving above equation we get m = 29, 30
M cannot be negative
Hence m = 30Sum of infinite terms of a geometric progression is 12. If the first term is 8, what is the 4th term of this GP?
 8/27
 4/27
 8/20
 1/3
S_{infinity} = a/(1  r )
12 = 8/(1r)
r = 1/3
Now, t_{4} = 8 x (1/3)^{3}
t_{4} = 8/27Find sum : 2^{2} + 2 x 3^{2} + 3 x 4^{2} + 4 x 5^{2} + ... + 10 x 11^{2}
 6530
 3600
 2850
 3850
Series is in the form n(n + 1)^{2} = n^{3} + 2n^{2} + n
Taking summation, Σ(n^{3} + 2n^{2} + n) = Σn^{3} + 2Σn^{2} + Σn Σ = [n(n+1)/2]^{2 }+ 2[n(n+1)(2n+1)/6] + n(n+1)/2
Substitute n = 10, required sum = 3850Σn