2IIM Quant Notes - Arithmetic and Geometric Progression


  • 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. Pn is the product of n terms of this GP. P6 > P5 and P6 > P7, what is the sum of all possible values of n?

    1. 4
    2. 9
    3. 5
    4. 13

    Common ratio is positive, and one of the terms is positive = > All terms are positive
    P6 = P5 * t6 = > If P6 > P5, t6 > 1
    P7 = P6 * t7 = > If P6 > P7, t7 < 1
    t6 = t2 * r4 = 1000r4;
    t7 = t2 * r5 = 1000r5
    1000r4 > 1 and 1000r5 < 1


    1/r4 < 1000 and 1/r5 > 1000.
    1/r = n.
    n4 < 1000 and n5 > 1000, where n is a natural number
    n4 < 1000 = > n < 6
    n5 > 1000 = > n ≥ 4
    n could be 4 or 5. Sum of possible values = 9 Correct Answer: 9

    Sum 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?

    1. 92
    2. -92
    3. 46
    4. 231

    Sum of first 12 terms is equal to sum of first 14 terms.
    Sum of first 14 terms = Sum of first 12 terms + 13th term + 14th term
    = > 13th term + 14th term = 0
    Let us assume 13th term = k, common ratio = r.
    14th 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: 92

    Sum of first 25 terms in arithmetic progression is 525, sum of the next 25 terms is 725, what is the common difference

    1. 8/25
    2. 4/25
    3. 6/25
    4. 1/25

    The sum of the first 25 terms, S25 = 525
    The sum of the next 25 terms, K25 = sum of next 25 terms = 725
    a26 = a1 + 25d
    a27 = a2 + 25d
    K25 = a26 + a27 + ...... + a50
    Or K25 = a1 + 25d + a2 + 25d + ...... + a25 + 25d
    Or K25 = a1 + a2 + ...... + a25 + 25(25d)
    Or K25 = S25 + 25(25d)
    i.e., 725 = 525 + 25 x 25d
    200 = 25 x 25d
    8 = 25d
    d = 8/25

    Let the nth term of arithmetic progression be defined as tn, and sum up to 'n' terms be defined as Sn. If |t8| = |t16| and t3 is not equal to t7, what is S23?

    1. 23(t16 - t8)
    2. 0
    3. 23t11
    4. Cannot be determined

    |t8|=|t16|. This can happen under two scenarios t8 = t16 or t8 = – t16.
    If t8 = t16, the common difference would be 0 suggesting that t3 would be equal to t7. However, we know t3 is not equal to t7, so the common difference cannot be zero.
    This tells us that t8 = – t16 Or, t8 + t16 = 0.
    If t8 + t16 = 0, then t12 = 0. t12 = t8 + 4d, and t16 – 4d So, t12 = ( t8 + t16 ) / 2
    For any two terms in an AP, the mean is the term right in between them. So, t12 is the arithmetic mean of t8 and t16.
    So, t12 = 0.
    Now, S23 = 23 × t12. 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, S23 = 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.

    1. i and ii only
    2. ii and iii only
    3. all three statements are true
    4. 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   b

    Statement (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 true

    Consider 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.?

    1. 40000
    2. 32000
    3. 8000
    4. 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: 40000

    If 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.?

    1. 7 times the 13th term
    2. 0
    3. 13 times the 7th term
    4. 4 times the 4th term + 9 times the 9th term

    4t 4 = 9t 9 , we need to find t13
    4(a + 3d) = 9(a + 8d)
    4a + 12d = 9a + 72d
    = > 5a + 60d = 0
    = > a + 12d = 0
    = > t13 = 0
    = > 13 × t"13" = 0
    As a simple rule, if n × tn = m × tm, then tm+n = 0. See, if you can prove this.
    Answer choice (b).
    Correct Answer: 0

    Sequence P is defined by pn = pn-1 + 3, p1 = 11, Sequence Q is defined as qn = qn-1 – 4, q3 = 103. If pk > qk+2, what is the smallest value k can take?

    1. 6
    2. 11
    3. 14
    4. 15

    Sequence P is an A.P. with a = 11, and common difference 3.
    So, Pk = 11 + (k – 1)3.
    Sequence Q is an A.P. with third term 103 and common difference – 4.
    t3 = a + 2d
    103 = a + 2 (– 4) or a = 111
    qk+2 = 111 + (k +1) (– 4)
    qk + 2 = 111 – 4k – 4 = 107 – 4k
    pk > qk + 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: 15

    The 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?

    1. 9
    2. 10
    3. 12
    4. 14

    First A.P., a = 1, d = 4
    S2n = 2n/2 [2 x 1+(2n −1)4]
    S2n = n(2 + 8n – 4) = n(8n – 2) = 8n2 – 2n
    For the second arithmetic progression, a = 56, d = 2
    Sn = n/2 [2 x 56+(n −1)2]
    Sn = 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, …}
    8n2 – 2n > n/2 (110+2n)
    16n2 – 4n > 110n + 2n2
    14n2 > 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?

    1. They are also in A.P.
    2. They are also in H.P.
    3. They are also in G.P.
    4. 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.

    1. 124
    2. 108
    3. 96
    4. 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 + (n-1)d]
    = > Sum upto 8 terms in this arithmetic progression = 8/2 * [2*5 + (8-1)3] = > 4*[10+21] = 4*31 = 124.

    If Sn = n3 + n2+ n+ 1 where Sn denotes the sum of the first n terms of a series and tm = 291, then m is equal to?

    1. 24
    2. 30
    3. 26
    4. 20

    The pattern is as follows : 2, 23, 234...
    Sn-Sn-1 = tn
    Substitute m instead of n
    Sm-Sm-1 = tm
    We know that Sn = n3 + n2+ n+ 1
    Hence m3+m2+m+1
    m3+m2+m+1-[(m-1)3+(m-1)2+(m-1)+1]=291
    m3+m2+m+1-[(m)3-(3m)2+3m-1+(m)2-2m+1+m-1+1]=291
    1+3m2+3m+1-2m-1=291
    -3m2+m-290=0
    3m2-m+290=0
    Solving above equation we get m = -29, 30
    M cannot be negative
    Hence m = 30

    Sum of infinite terms of a geometric progression is 12. If the first term is 8, what is the 4th term of this GP?

    1. 8/27
    2. 4/27
    3. 8/20
    4. 1/3

    Sinfinity = a/(1 - r )
    12 = 8/(1-r)
    r = 1/3
    Now, t4 = 8 x (1/3)3
    t4 = 8/27

    Find sum : 22 + 2 x 32 + 3 x 42 + 4 x 52 + ... + 10 x 112

    1. 6530
    2. 3600
    3. 2850
    4. 3850

    Series is in the form n(n + 1)2 = n3 + 2n2 + n

    Taking summation, Σ(n3 + 2n2 + n) = Σn3 + 2Σn2 + Σn Σ = [n(n+1)/2]2 + 2[n(n+1)(2n+1)/6] + n(n+1)/2

    Substitute n = 10, required sum = 3850Σn


Log in to reply
 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.