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Class 11 Applied Maths Chapter 6 Solutions
Sequences and Series
EXERCISE- 6.4
Q.1 (i) Find the 15th term of the series √3 + 1/√3 + 3√3 +…..
(ii) Find the 10th and the nth terms of the sequence 5, 25, 125, …
(iii) Find the 6th term from the end of the sequence 3, -6, 12,-24, …, 12288.
Ans. (i) a = √3, r = (1/√3)/√3 = 1/(√3x√3) = 1/3
a15 = a.r^(15-1)
= a.r^16
= (√3)(1/3)^14
= (3)^1/2 (3)^-14
= 3^(1/2-14)
= 3^(-27/2)
(ii) a =5, r = 5, a10 = ?, an = ?
Using, an = a x r^(n-1), we get,
an = 5 x 5^(n-1)
= 5^(1+n-1) = 5^n
Now, a10 = 5^10
(iii) a = 3, r = 5, a6 formed = ?
a6 formed = an x (1/r)^(n-1)
= 12288 x (1/-2)^5
= 12288 x 1/-32
= -384
Q.2 Which term of the sequence
(i) 2, 2√2, 4, … is 128?
(ii) 1, 1/3, 1/9 ,…is 1/243?
Ans. (i) a = 2, r = √2, an = 128, n = ?
Using ,an x r^(n-1), we get,
128 = 2 x (√2)^(n-1)
128/2 = [2^(1/2)]^(n-1)
64 = 2^[(n-1)/2]
2^6 = 2^[(n-1)/2]
Therefore, 6 = (n-1)/2
12 = n – 1
n = 13th
(ii) a = 1, r = 1/3, an = 1/243, n = ?
Using ,an x r^(n-1), we get,
1/243 = 1 x (1/3)^(n-1)
1/3^5 = 1/[3]^(n-1)
3^(n-1) = 3^5
n – 1 = 5
n = 6th
Q.3 Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.
Ans. ATQ, A8 = 192, r = 2
Using, An = a x r^(n-1), we get,
a x r^7 = 192
a x 2^7 = 192
a x 128 = 192
a = 192/128
a = (2^6×3)/2^7
a = 3/2
Now, A12 = a x r^11
= 3/2 x 2^11
= 3 x 2^10
= 3 x 1024
= 3072
Q.4 In a G.P., the third term is 24 and 6th term is 192. Find the 10th term.
Ans. A3 = 24, A6 = 192
Using An = a x r^(n-1), we get,
A3 = ar^2 = 24……..(i)
A6 = ar^5 = 192……(ii)
Dividing (ii) by (i)
ar^5/ar^2 = 192/24
r^3 = 8
r = 2
Substituting r = 2 in eq. (i), we get,
a x 2^2 = 24
a = 24/4
a = 6
Now, a10 = a x r^9
= 6 x 2^9
= 6 x 512
= 3072
Q.5 Find the number of terms of a G.P. whose first term is is 3/4,common ratio is 2 and the last term is 384.
Ans. ATQ, A1 = 3/4, r = 2, An = 384, n = ?
Using An = A x r^(n-1), we get,
384 = 3/4 x 2^(n-1)
384 x 4/3 = 2^(n-1)
2^7 x 2^2 = 2^(n-1)
2^9 = 2^(n-1)
Therefore, 9 = n – 1
n = 10
Q.6 Find the value of x such that
(i) -2/7, x, -7/2 are the three consecutive terms of a G.P are three consecutive terms of a G.P.
(ii) x+9, x-6 and 4 are three consecutive terms of a G.P.
Ans. (i) ATQ, -2/7, x, -7/2…..(GP)
b^2 = a x c
x^2 = -2/7 x -7/2
x^2 = 1
x = ±1
(ii) ATQ, x+9, x-6, 4……..(GP)
b^2 = ac
(x-6)^2 = (x+9)4
x^2 + 36 – 12x = 4x + 36
x^2 = 12x + 4x
x^2 = 16x
x = 16
Q.7 The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q^2 = ps.
Ans. ATQ, A5 = p = ar^4
A8 = q = ar^7
A11 = s = ar^10
Using, An = a + r^(n-1), we get,
L.H.S :
q^2 = [ar^7]^2 = a^2r^14
R.H.S :
ps = ar^4 x ar^10 = a^2r^14
LHS = RHS
Hence proved.
Q.8 If a, b, c are in A.P., then show that 3^a, 3^b, 3^c are in G.P.
Ans. ATQ, a, b, c….(AP)
b – a = c – b = D……….(i)
Now, we have,
A1 = 3^a, A2 = 3^b, A3 = 3^c
For A1, A2& A3 to be in G.P
A2/A1 = A3/A2
A2/A1 = 3^b/3^a = 3^(b-a) = 3^D…….[from (i)]
A3/A2 = 3^c/3^b = 3^(c-b)= 3^D………[from (i)]
Since, A2/A1 = A3/A2 = D
Therefore, A1, A2, A3………..(GP)
Q.9 If k, 2k +2, 3k+3,… are in G.P., then find the common ratio of the G.P.
Ans. ATQ, k, 2k + 2, 3k + 3,……….(GP)
Now, we know that in a GP,
(A2)^2 = A1 x A3
(2k – 2)^2 = k(3k + 3)
4k^2 + 4 + 8k = 3k^2 + 3k
4k^2 – 3k^2 + 8k – 3k + 4 = 0
k^2 + 5k + 4 = 0
k^2 + 4k + k + 4 = 0
k(k + 4) + 1(k + 4) = 0
(k + 4 ) (k + 1 ) = 0
k = -4 or -1
If k = -4, we get,
-4, -6, -9,…….
r = 3/2
If k = -1, we get,
-1, 0, 0
Therefore, r = 3/2
Q.10 If a^x = b^y = c^z such that a, b, c are in G.P. and x, y, z are unequal positive integers, then show that 2/y = 1/x + 1/z.
Ans. ATQ, a, b, c…….(GP)
Also, a^x = b^y = c^z
Let, a^x = b^y = c^z = k
Therefore, a = k^(1/x)
b = k^(1/y)
c = k^(1/z)
Since, a, b, c……(GP)
b^2 = ac
[k^(1/x)]^2 = k^(1/x) x k^(1/z)
k^(2/y) = k^(1/x+1/z)
2/y = 1/x + 1/z
Q.11 The sum of three consecutive terms of a G.P. is 26 and their product is 216. Find the common ratio and the terms of the G.P.
Ans. Let, three term of GP be a/r, a, ar.
ATQ, a/r x a x ar = 216
a^3 = 216
a = 6
Now, three terms would be, 6/r, 6, 6r
Also, 6/r + 6 + 6r = 26
6/r + 6r = 26 – 6
6[1/r + r] = 20
(1 + r^2)/r = 20/6
3(1 + r^2) = 10r
3 + 3r^2 = 10r
3r^2 – 10 r + 3 = 0
3r^2 – 9r – r + 3 = 0
3r(r – 3) -1 (r – 3) = 0
(r – 3) (3r – 1) = 0
r = 3 or 1/3
If r = 3, we get,
GP : 2, 6, 18
If r = 1/3, we get,
GP : 18, 6, 2
Q.12 Find four numbers forming a G.P. in which third term is greater than the first term by 9 and second term is greater than the fourth term by 18.
Ans. a3 = a + 9, a2 = a4 + 18
ar^2 = a + 9 ar = ar^3 + 18
ar^2 – a = 9 ar -ar^3 = 18
a(r^2 + 1) = 9 -ra(-1 + r^2) = 18
putting value of a(r^2 + 1) = 9
In, -ra(r^2 + 1) = 18
-r x 9 = 18
r = -2
Therefore, a(r^ – 1) = 9
a(4 – 1) = 9
a = 9 – 3 = 6
So, terms will be :
3, 3 x -2, 3 x (-2)^2, 3 x (-2)^3
3, -6, 12, -24
Q.13 Three positive numbers form an increasing G.P. If the middle term of the series is doubled, then the new numbers are in A.P. Find the common ratio of the G.P.
Ans. a, ar, ar^2 is a GP
After doubling the middle term AP will be,
a, 2ar, ar^2 is an AP
2(2ar) = a + ar^2
4ar = a(1 + r^2)
0 = 1 + r^2 – 4r
r^2 – 4r + 1 = 0
r = -b±√b^2-4ac/2a
= -4±√16-4(1)(1)/2×1
= 4±√12/2
= 4±2√3/2
= 2(2±√3)/2
= 2 ± √3
Q.14 The fourth term of a G.P. is the square of its second term and the first term is -3. Determine its seventh term.
Ans. ATQ, A1 = -3 (a) , A4 = (A2)^2
Using, An = a x r^(n-1), we get,
ar^3 = (ar)^2
ar^3 = a^2r^2
r = a
Therefore, r = -3
Now, a7 = a x r^6
= -3 x -3^6
= -3^7
= -2187
Q.15 The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Ans. Let, three no. in GP be a/r, a, ar
ATQ, a/r x a x ar = 1
a^3 = 1
a = 1
Now, our three no. would be,
1/r, 1, r
1/r + 1 + r = 39/10
1/r + r = 39/10 – 1/1
(1 + r^2)/r = (39 – 10)/10
(1 + r^2)/r = 29/10
10(1 + r^2) = 29r
10 + 10r^2 – 29r = 0
10r^2 – 29r + 10 = 0
10r^2 – 25r – 4r + 10 = 0
5r(2r – 5) – 2(2r – 5) = 0
(2r – 5) (5r – 2) = 0
r = 5/2 or 2/5
If r = 5/2, we get,
2/5, 1, 5/2
If r = 2/5, we get,
5/2, 1, 2/5
Q.16 Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.
Ans. Let, three no. in AP be, a – d, a , a + d
ATQ, a – d + a + a + d = 15
3a = 15
a = 5
Now, three terms would be : 5-d, 5, 5+d
Also ATQ, (5-d+1), (5-4), (5+d+19)
(6-d), 9, (24+d)…….[GP]
We know that, in a GP
(9)^2 = (6-d) (24+d)
81 = 144 – 24d + 6d – d^2
81 – 144 = -18d – d^2
-63 + d^2 + 18d = 0
d^2 + 18d – 63 = 0
d^2 + 21d -3d -63 = 0
d(d + 21) – 3(d + 21) = 0
(d + 21 ) ( d – 3) = 0
d = -21 or 3
If, a = 5, d = -21
Then, -5(-21), 5, (5 – 21)
26, 5, -16
If, a = 5, d = 3
Then, (5 – 3), 5, (5 + 3)
2, 5, 8
Q.17 If a, b, c are in G.P., prove that the following are also in G.P.
(i) a^3, b^3, c^3
(ii) a^2+ b^2, ab + bc, b^2 + c^2.
Ans. (i) a, b, c are in AP
b^2 = ac
(b^2)^3 = (ac)^3
(b^3)^2 = a^3 c^3
Therefore, a^3, b^3, c^3 are in AP
(ii) a^2+ b^2, ab + bc, b^2 + c^2
-(ab + bc)^2 = [b(a + c)]^2
= b^2 (a + c)^2
= ac (a + c)^2
(a^2 + b^2) (b^2 + c^2) = (a^2 + ac) (ac + c^2)
= a(a + c) c(a + c)
= ac (a + c)^2
Q.18 If a, b, c, d are in G.P., show that
(i) a^2 + b^2, b^2 + c^2, c^2 + d^2 are in G.P.
(ii) (b-c)^2+ (c-a)^2 + (d-b)^2= (a-d)^2.
Ans. (i) A1 = a, A2 = b = ar, A3 = c = ar^2, A4 = d = ar^3
For, a^ + b^2, b^2 + c^2, c^2 + d^2……….GP
(b^2+c^2)/(a^2+b^2) = (c^2+d^2)/(b^2+c^2) = CR
Now, (b^2+c^2)/(a^2+b^2) = [(ar)^2+(ar^2)^2]/[a^2+(ar)^2] = [a^2r^2+a^2r^4]/[a^2+a^2r^2] = a^2r^2[1+r^]/a^2[1+r^2] = r^2
(c^2+d^2)/(b^2+c^2) = [(ar^2)^2+(ar^3)^2]/[(ar)^2+(ar^2)^2] = (a^2r^4 a^2r^6)/(a^2r^2)+(a^2r^4) = a^2r^4(1+r^2)/a^2r^2(1+r^2) = r^2
Since, (b^2 + c^2)/(a^2 + b^2) = (c^2 + d^2)/(b^2 + a^2) = r^2(CR)
Therefore, Terms are in AP.
(ii) A1 = a, A2 = b = ar, A3 = c = ar^2, A4 = d = ar^3
L.H.S
(ar-ar^2)^2 + (ar^2-a)^2 + (ar^3-ar)^2
(ar)^2 + (ar^2)^2 – 2 x ar x ar^2 + (ar^2)^2 + a^2 – 2 x ar^2 x ar + (ar^3)^2 + (ar)^2 – 2 x ar^3 x ar
a^2r^2 + a^2r^4 – 2a^2r^3 + a^2r^4 + a^2 – 2a^2r^2 + a^2r^6 + a^2r^2 – 2a^2r^4
a^2 + a^2r^6 – 2a^2r^3
(a)^2 + (ar^3)^2 – 2 x a x ar^3
a^2 + d^2 – 2ad – (a – d)^2 = R.H.S
Q.19 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?
Ans. ATQ, A0 = 30
A1 = 30 x 2^1 = 60
A2 = 30 x 2^2 = 120
A4 = 30 x 2^4 = 30 x 16 = 480
An = 30 x 2^n
Q.20 The lengths of the sides of a triangle form a G.P. If the perimeter of the triangle is 37 cm and the shortest side is of length 9 cm, find the lengths of the other two sides.
Ans. Let, the length of the side of triangle be a/r, a, ar
ATQ, Perimeter = 37
Since, Shortest side is 9 cm
a/r = 9 cm
a = 9r
Now, a/r + a + ar = 37
9 + a + ar = 37
a + r = 37 – 9
a + r = 28
9r + 9r^2 = 28
9r^2 + 9r – 28 = 0
9r^2 – 12r + 21r – 28 = 0
3r(3r – 4) + 7(3r – 4) = 0
(3r – 4) (3r + 7) = 0
r = 4/3 or -7/3 (Rejected)
Therefore, r = 4/3
a = 9 x 4/3 = 12 cm
ar = 12 x 4/3 = 16 cm
Q.21 An insect population is growing in such a way that each generation is 2.5 times as large as the previous one. If there are 10000 insects in the first generation, how many are there in the 5th generation?
Ans. A1 = 10000
A2 = 10000 x (2.5)^1
.
.
A5 = 10000 x (2.5)^4 = (25)^4
Q.22 The population of a country is 114 million at present. If the population is growing at the rate of 10% per year, then what will be the population of the country in 5 years?
Ans. A1 = 114 M
A2 = 114 x (1.1)^1 M
A5 = 114 x (1.1)^4 M
Q.23 A manufacturer reckons that the value of a machine, which costs him 56200 will depreciate each year by 20%. Find the estimated value at the end of 3 years.
Ans. Value of machine at the beginning of 1st year = 56200
Depreciation is applied for 3 years
Value of machine at the end of year 3 = 56200 x [80/100]^3
= 56200 x [4/5]^3
= 56200 x 64/125
= Rs.28745 (Approx)
Q.24 Priyanka invested Rs.1300 in an account that pays 4% interest compounded annually. Assuming no deposits or withdrawals are made, find how much money she would have in the account 6 years after her initial investment?
Ans. A = P[1 + R/100]^n
= 1300[1 + 4/100]^6
= 1300 (1 + 0.04)^6
= 1300 (1.04)^6
FAQ’s related to Class 11 Applied Maths Chapter 6 on Sequences and Series:
Q.1 What is a sequence?
Ans. A sequence is an ordered list of numbers following a particular pattern. Each number in a sequence is called a term.
Q.2 What is a series?
Ans. A series is the sum of the terms of a sequence. If we add the terms of a sequence, we get a series.
Q.3 What is the difference between an arithmetic sequence and a geometric sequence?
Ans.
- Geometric Sequence – A geometric sequence (or geometric progression) is a sequence in which each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio (r).
- Arithmetic Sequence – An arithmetic sequence (or arithmetic progression) is a sequence in which the difference between consecutive terms is constant. This difference is called the common difference (d).
These are a few Frequently Asked Questions relating to Class 11 Applied Maths Chapter 6
In Class 11 Applied Maths chapter 6, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 11 Applied Maths Chapter 6 promises an enriching learning experience that will set you on the path to success. Class 11 Applied Maths Chapter 6, we delve deep into advanced mathematical concepts that are crucial for understanding.
Class 11 Applied Maths Chapter 6 Exercise :
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