Sequences and Series – Exercise 9.3[21-32] – Class XI

  1. Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Solution:

Let a be the first term and r be the common ratio of the G.P.

a1 = a, a2 = ar, a3 = ar2, a4 = ar3

By the given condition,

a3 = a1 + 9 ⇒ ar2 = a + 9 ………………. (1)

a2 = a4 + 18 ⇒ ar = ar3 + 18 ………… (2)

From (1) and (2), we obtain

a(r2 – 1) = 9 …………………….. (3)

ar (1– r2) = 18 …………………. (4)

Dividing (4) by (3), we obtain

[ar(1-r2)]/[a(r2-1)] = 18/9

-r = 2

r = -2

Substituting the value of r in (1), we obtain

4a = a + 9

⇒ 3a = 9

∴ a = 3

Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3 i.e., 3 ̧–6, 12, and –24.


  1. If pth, qth and rth terms of a G.P. are a, b and c respectively. Prove that aq-r. br-p.Cp – q = 1Solution:Let A be the first term and R be the common ratio of the G.P.

    According to the given information,

    ARp-1 = a

    ARq-1 = b

    ARr-1 = c

    aq-r. br-p.Cp – q = Aq-r x R(p-1)(q-r) x Ar-p x R(q-1)(r-p) x Ap – q x R(r-1)(p-q)

    =Aq-r+r-p+p-q x R(pr – pr – q –pq) + (rq – r + p – pq) + (pr – p – qr + q)

    = A0xR0

    = 1

    Thus, the given result is proved.


    1. If the first and the nth term of G.P. are a and b, respectively, and if p is the product of n terms, prove that p2 = (ab)n

    Solution:

    The first term of the G.P is a and the last term is b.

    Therefore, the G.P. is a, ar, ar2, ar3 … arn–1, where r is the common ratio.

    b = arn–1 ……………………………. (1)

    P = Product of n terms

    = (a) (ar) (ar2) … (arn–1)

    = (a × a ×…a) (r × r2 × …rn–1)

    = an r1 + 2 +…(n–1) … (2)

    Here, 1, 2, …(n – 1) is an A.P.

    1 + 2 + ……….+ (n – 1) = (n-1)/2 [2 + (n-1-1) x 1] = (n-1)/2 [2 + n – 2] = n(n-1)/2=

    P = anr[n(n-1)/2]

    P2 = a2n. rn(n-1)

    = [a2.r(n-1)]n

    =[ a x arn-1]n

    = (ab)n

    Thus, the given result is proved.


    1. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rn

    Solution:

    Let a be the first term and r be the common ratio of the G.P.

    Sum o first n terms = [a(1-rn)]/(1-r)

    Since there are n terms from (n+1)th to (2n)th term,

    Sum of terms from (n +1)th to (2n)th term

    1Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rn.


    1. If a, b, c and d are in G.P. show that (a2 + b2 + c2)( b2 + c2 + d2) = (ab + bc – cd)2

    Solution:

    a, b, c, d are in G.P. Therefore,

    bc = ad …………………… (1)

    b2 = ac …………………… (2)

    c2 = bd ………………….. (3)

    It has to be proved that,

    (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2

    R.H.S.

    = (ab + bc + cd)2

    = (ab + ad + cd)2 [Using (1)]

    = [ab + d (a + c)]2

    = a2b2 + 2abd (a + c) + d2 (a + c)2

    = a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)

    = a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)]

    = a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2

    = a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2

    [Using (2) and (3) and rearranging terms]

    = a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)

    = (a2 + b2 + c2) (b2 + c2 + d2) = L.H.S.

    ∴ L.H.S. = R.H.S.

    ∴ (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2 


    1. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

    Solution:

    Let G1 and G2 be two numbers between 3 and 81 such that the series, 3,

    G1, G2, 81, forms a G.P.

    Let a be the first term and r be the common ratio of the G.P.

    ∴ 81 = (3) (r)3

    ⇒ r3 = 27

    ∴ r = 3 (Taking real roots only)

    For  r = 3,

    G1 = ar = (3) (3) = 9

    G2 = ar2 = (3) (3)2 = 27

    Thus, the required two numbers are 9 and 27.


    1. Find the value of n so that [an+1+bn+1]/[an+bn] may be the geometric mean between a and b.

    Solution:

    1. of a and b is √ab

    By the given condition [an+1+bn+1]/[an+bn] = √ab

    Squaring both sides, we obtain

    [an+1+bn+1]2/[an+bn]2 = ab

    a2n+2 + 2an+1bn+1 + b2n+2 = (ab)(a2n +2anbn + b2n)

    a2n+2 + 2an+1bn+1 + b2n+2 = a2n+1b +2an+1bn+1 + ab2n+1

    a2n+2 + b2n+2 = a2n+1b+ab2n+1

    a2n+2 – a2n+1b = ab2n+1 – b2n+2

    a2n+1(a-b) = b2n+1(a-b)

    (a/b)2n+1 = 1 = (a/b)0

    2n + 1 = 0

    n = –1/2


    1. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2)( 3 – 2√2)

    Solution:

    Let the two numbers be a and b.

    G.M. = √(ab)

    According to the given condition,

    a + b = 6√(ab) ———(1)

    (a+b)2 = 36(ab)

    Also, (a –b)2 = (a + b)2 – 4ab = 36ab – 4ab = 32 ab

    a – b = √32. √(ab)

    =4√2. √(ab) ———(2)

    Adding (1) and (2) , we get

    2a = (6 + 4√2) √(ab)

    a = (3 + 2√2) √(ab)

    Substituing the value of a in (1), we get

    b = 6√(ab) – (3+2√2) √(ab)

    b = (3 – 2√2) √(ab)

    a/b = [(3 + 2√2) √(ab)]/[ (3 – 2√2) √(ab)] = 3 + 2√2/3 + 2√2

    Thus, the required ratio is (3 + 2√2):(3 – 2√2)


    1. If A and G be A.M. and G.M. respectively between two positive numbers prove that the numbers are A ± √(A+G)(A-G)

    Solution:

    It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.

    A.M. = A = a+b/2 ————–(1)

    GM = G = √(ab) ——————-(2)

    From (1) and (2), we get

    a + b = 2A ————-(3)

    ab = G2 —————(4)

    Substituting the value of a and b from (3) and (4) I the identity

    (a – b)2 = (a + b)2 – 4ab, we get

    (a – b)2 = 4A2 – 4G2 = 4(A2 – G2)

    (a – b)2 = 4(A+G)(A-G) ————-(5)

    From (3) and (5), we get

    2a = 2A + 2√[(A+G)(A – G)]

    a = A + √[(A+G)(A – G)]

    substituting the value of a in (3), we obtain

    b = 2A – A – √(A+G)(A-G) = A – √(A+G)(A-G)

    Thus, the two numbers are A ± √(A+G)(A-G)


    1. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

    Solution:

    It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P.

    Here, a = 30 and r = 2

    a3 = ar2 = (30) (2)2 = 120

    Therefore, the number of bacteria at the end of 2nd hour will be 120.

    a5 = ar4 = (30) (2)4 = 480

    The number of bacteria at the end of 4th hour will be 480.

    an +1 = arn = (30)2n

    Thus, number of bacteria at the end of nth hour will be 30(2)n.


    1. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

    Solution:

    The amount deposited in the bank is Rs 500.

    At the end of first year, amount = Rs 500 (1+1/10) = Rs. 500(1.1)

    At the end of 2nd year, amount = Rs 500 (1.1) (1.1)

    At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on

    Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

    = Rs 500(1.1)10


    1. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

    Solution:

    Let the root of the quadratic equation be a and b.

    According to the given condition,

    A.M. = a+b/2 = 8

    a + b = 16 ——–(1)

    G.M. = √(ab) = 5

    ab = 25 ————-(2)

    The quadratic equation is given by,

    x2– x (Sum of roots) + (Product of roots) = 0

    x2 – x (a + b) + (ab) = 0

    x2 – 16x + 25 = 0 [Using (1) and (2)]

    Thus, the required quadratic equation is x2 – 16x + 25 = 0


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