Progressions - Vikas Saini



  • Arithmetic Progression

    Basic formulae :-

    Tn = a + (n-1)d.

    Sn= (n/2) [2a+(n-1)d].

    S- Sn-1 = d.

    Here Tn = nth term.

    Sn = Sum of n terms.

    a = first term.

    d = difference.

    Q. Sum of 6 terms of an  AP is 6. 3rd term of AP is 0.

    Find the 6th term of it.

    Solution :-

    Suppose 3rd term of AP is a.

    2nd & 1st term are a-d & a-2d respectively.

    4th, 5th & 6th terms are a+d,a+2d,a+3d respectively.

    So (a-d)+(a-2d)+a+(a+d)+(a+2d)+(a+3d)  = 6.

    = > 6a + 3d = 6.

    = > 0 + 3d = 6.

    = > d = 2.

    So a+3d = 6.

    6th term is 6.

    Q. Sum of 3 terms of an AP is 30. If we add 6 in product of these 3 terms then it becomes a perfect square. Find all 3 terms of AP.

    Solution :-

    Let three terms are a-d, a, a+d.

    = > (a-d)+a+(a+d) = 30.

    = > 3a = 30.

    = > a = 10.

    (10 – d) x 10 x (10+d) + 6 = k^2.

    = > 10(100 – d^2)+6 = k^2.

    d = 1,2,3,4,5,6,7,8,9.

    d^2 = 1,4,9,16,25,36,49,64,81.

    100 – d^2 = 99,96,91,84,75,64,51,36,19.

    10(100 – d^2)+6 = 996,966,916,846,756,646,516,366,196.

    196 is the only number here which is perfect square.

    So d = 9.

    3 terms of AP are = (10-9),10,(10+9)

    = 1,9,19.

    Q. Sum of first 4 terms of an AP is 42. Sum of square of 4 terms are 566.Find all the 4 terms.

    Solution :-

    Let terms are a-2d, a-d , a+d , a+2d.

    (a-3d)+(a-d)+(a+d)+(a+3d) = 42.

    = > 4 a  = 42.

    a = 10.5.

    (a-3d)^2 + (a-d)^2 + (a+d)^2  + (a+3d)^2 = 566.

    = > a^2 + 9d^2 – 6ad + a^2 + d^2 – 2ad + a^2 + d^2 + 2ad + a^2 + 9d^2 + 6ad  = 566.

    = > 4a^2 + 20 d^2 = 566.

    = > 4 x (10.5)^2 + 20 d^2 = 566.

    = > 441 + 20d^2 = 566.

    =  > 20 d^2 = 125.

    = > d^2 = 6.25

    = > d = 2.5 .

    Numbers are 3,8,13,18.

    Q. From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term ?

    Solution :-

    For d=1, total = 16.

    (1,2,3,4,5)(2,3,4,5,6)………………………..(16,17,18,19,20)

    For d=2, total = 6.

    (2,4,6,8,10)(4,6,8,10,12)…….(12,14,16,18,20).

    For d=3, total = 2.

    (3,6,9,12,15)(6,9,12,15,18 )

    For d=4, total = 1.

    (4,8,12,16,20)

    Total = 16+6+2+1 = 25.

    Q. Ratio of sum of two APs are (n+1 : 2n).Find ratio of 6th terms of both series.

    Solution :-

    To find ratio of any term m.

    We need to put n = 2m – 1.

    m = middle term.

    Here m = 6.

    n = 11.

    So ratio of 6th terms = 11 + 1 / 2 x 11 = 12 / 2 x  11

    = 6 / 11.

    Q. Find total no of APs with 5 distinct terms that can be formed from the first 50 natural numbers.

    Solution :-

    For d = 1, Total 46.

    (1,2,3,4,5),(2,3,4,5,6)…………………….(46,47,48,49,50)

    For d=2, total 42.

    (1,3,5,7,9),(2,4,6,8,10)…………………(42,44,46,48,50)

    For d=3, total 38.

    (1,4,7,10,13),(2,5,8,11,14)………….(38,41,44,47,50)

    For d= 12, total 2.

    (1,13,25,37,49)(2,14,26,38,50).

    Total = 46+42+38…………….2.

    Total = (12/2) (2+46) = 288.

    Q. If log32 + log3(2x – 5) and log3(2x – 7/2) are in AP, then find the value of x.

    Solution n:-

    All nos are in AP.

    2 log3(2x – 5) = log32 + log3(2x – 7/2)

    Suppose 2x  = t.

    2 log3(t – 5) = log32 + log3(t – 7/2)

    = > log3(t-5)2 = log3(2t – 7)

    = > (t – 5 )2 = 2t – 7.

    = > t2 – 10t + 25 = 2t – 7

    = > t2 – 12t + 32 = 0.

    = > t = 4,8

    2x = 4 , 8

    x = 2,3.

    Q. A sequence of terms is defined such that

    2an  = an+1 + an-1;

    a0 = 1; a1 = 3.

    What is the value of a0 + a1 + a2 ... a50.

    Solution :-

    It shows property of an AP.

    a0 =1, a1 = 3.

    d = 2.

    a50 = a0 + 50 d = 1 + 50 x 2 = 101.

    Sum = (51/2) (1+101) = 51 x 51 = 2601.

    Geometric Progression

    Tn  = arn-1

    Sn = a (rn – 1) / (r – 1) if r > 0

         = a ( 1 – rn )/ (1 – r ) if r < 0.

    Sum of infinite terms = a / (1 – r).

    Q. Tn is the nth term of geometric progression such that Tn = Tn-1 + Tn-2 for n > 2. If the first term and common ratio of the progression are positive real numbers, then find the common ratio.

    Solution :-

    T3 = T2 + T1.

    ar2 = ar + a.

    r2 = r + 1.

    = > r2 – r – 1 = 0.

    By using Shridhar acharya’s method

    r = 1+- root (1+4) / 2

    r = (1+ root 5 )/ 2.

    Q. If the roots of the equation ax3+bx2+cx+d = 0 are in geometric Progression, then which of the following relation is true?

    (a) ac2 = bd                             (b) ac3=b3d

    (c) a2c = bd2                            (d) a3c=bd3

    Solution :-

    Let roots are 1,2 & 4.

    a = 1.

    b = -(1+2+4) = -7.

    c = 1 x 2 + 2 x 4 +1 x 4 = 2+8+4=14.

    d = -(1 x 2 x 4) = - 8.

    (c / b)3 = -8.

    d / a = -8.

    = > (c/b)3 = d/a.

    = > ac3  = b3d.

    Harmonic Progression

    Q. ‘M’ and ‘N’ are natural numbers such that by M = (5N – 4)(5N + 1).

    If 1 < = N < = 200, what is the harmonic mean of all possible values of M ?

    Solution :-

    N =1, M = 1 x 6.

    N = 2, M = 6 x 11.

    N = 200, M =996 x 1001.

    Total numbers / Harmonic mean = 1 / (1 x 6) + 1 / (6 x 11) ………………………….. 1/996 x 1001.

    = > 200 / Harmonic mean  = (1 / 5) ( 1 – 1/6 + 1/6 – 1/11…………… -1/1001)

    = > 200 / Harmonic mean = (1/5) ( 1 – 1/1001)

    = > 200 / Harmonic mean = (1 / 5) (1000 / 1001)

    = > Harmonic mean = 1001.

    Q. A person leaves his home at 6 AM at speed of 20 km/hr and reaches his office at 9 AM. When his speed is 30 km/hr, he reaches at 8 AM.

    If he has reach at 8:30 AM, what should be his speed?

    Solution :-

    Here we see time is in AP.

    So Speed must be in HP.

    We need to find harmonic mean of speed 20 & 30.

    Speed = 2 x 20 x 30 / (20 + 30) = 24 km/hr.

    Method – 2

    D = 20 x 3 = 30 x 2 = 60.

    Time = 2.5 hr.

    Speed = 60 / 2.5 = 24 km/hr.

    Miscellaneous Problems

    Q. If S = 12  - 22 + 32 - 42 ... + 20012.

    Then what is the value of S.

    Solution :-

    Method 1

    S = (12-22)+(32-42)……………………………….(19992 – 20002) + 20012

    S = (-3)+(-7)………………………………..(-3999)+20012.

    Total terms from (-3 to -3999) = (3999 – 3)/4 + 1 = 1000.

    S = (1000 / 2) (-3-3999) + 20012

    S = -2001000 + 4004001 = 2003001.

    Method 2

    S = (12+22+32………………..20012) – 2(22+42+62…………………….20002)

    S = (12+22…………………20012) – 8(12+22…………..10002)

    S  = (2001 x 4003 x 2002 / 6) – 8(1000 x 2001 x 1001 / 6)

    S = 2003001.

    Q. a1 = 1, a2 = 2 and an+2 = an(an+1 – 1),what is the value a1000 ?

    Solution :-

    a3 = a1(a2-1).

    a3=1(2 – 1) = 1.

    a4=2(1 – 1) = 0.

    a5= 1(0 – 1) = -1.

    a6 =0(10-1)= 0.

    a7 = -1(0-1)=1.

    a8=0.

    a9 = -1.

    a10 = 0.

    We see, after a1 and a2  cyclicity of 4 in rest terms.

    a998 = 0.

    a999 = 1.

    a1000 = 0.

    Q. 4x + 2y + z =35, where ‘x’,’y’ and ‘z’ are positive real numbers.

    Find maximum value of  x1/2  x y1/4 x z1/8.

    Solution :-

    4x = k/2.

    2y = k/4.

    z = k/8.

    (k/2)+(k/4)+(k/8 )  = 35.

    = > 4k+2k+k/8 = 35.

    = > 7 k/8  = 35

    = > k = 40.

    x = 5, y = 5, z=5.

    Max value = 57/8

     


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