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Class 11 Applied Maths Chapter 6 Solutions
Sequences and Series
EXERCISE- 6.5
Q.1 Find the sum of :
(i) 20 terms of the sequence 0.15, 0.015, 0.0015,…
(ii) 5 terms and n terms of the series 1 + 2/3 + 4/9 +…..
(iii) 10 terms and n terms of the series 3 + 6 + 12 +….
(iv) 8 terms and p terms of the series 1 – 3 + 9 – 27 +……
(v) 6 terms of the series √3 + 3 + 3√3 +…
(vi) n terms of the sequence √7, √21, 3√7,…
(vii) n terms of the sequence x^3, x^5, x^7,… (x ≠ ±1)
(viii) infinite terms of the series 7 – 1 + 1/7 – 1/49 +….
(ix) infinite terms of the series √2 – 1/√2 + 1/2√2 – 1/4√2 +…..
Ans. (i) a = 0.15, r = 0.015/0.15 = 0.1
Sn = a(1 – r^n)/(1 – r)
S20 = 0.15(1 – (0.1)^20)/(1 – 0.1)
= 0.15/0.9 [1 – (1/10)^20]
= 1/6 (1 – 1/10^20)
(ii) Here, a = 1, r = 2/3, n = 5,
Using, Sn = a(1 – r^n)/(1 – r), we get,
S5 = 1[1 – (2/3)^5]/[1 – (2/3)]
= [1 – (32/243)]/(1/3)
= (243 – 32)/243 x 3/1
= 211/81
Also, Sn = 1[1 – (2/3)^n]
= [1 – (2^n/3^n)]/(1/3)
= [1 – (2/3)^n] x 3/1
= 3 [1 – (2/3)^n]
(iii) a = 3, r = 6/3 = 2
Sn = a(r^n – 1)/(r – 1)
S10 = 3(2^10 – 1)/(2 – 1)
= 3(2^10 – 1)/1
= 3(1024 – 1)
= 3 x 1023
= 3069
Also, Sn = 3(2^n – 1)/(2 – 1)
= 3(2^n – 1)
(iv) Here, a = 1, r = -3, n = 8,
Using, Sn = a(r^n – 1)/(r – 1)
S8 = 1(-3^8 – 1)/(-3 – 1)
= (3^8 – 1)/-4
= (6561 – 1)/-4
= 6560/-4
= -1640
Also, Sp = 1((-3)^p – 1)/(-3 – 1)
= [(-3)^p – 1]/-4
= 1/4[1 – (-3)^p]
(v) Here, a = √3, r = √3, n = 6
Using, Sn = a(r^n – 1)/(r – 1)
S6 = √3((√3)^6 – 1)/(√3 – 1)
= √3(27 – 1)/(√3 – 1)
= √3(26)/(√3 – 1)
= 26√3/(√3-1) x (√3+1)/(√3+1)
= [26√3(√3+1)]/[(√3)^2 – (1)^2]
= [26√3(√3+1)]/(3 -1)
= [26√3(√3+1)]/2
= 13√3 x √3 + 13√3
= 13 x 3 + 13√3
= 39 + 13√3
(vi) Here, a = √7, r = √3
Using, Sn = a(r^n – 1)/(r – 1)
Sn = √7((√3)^n – 1)/(√3 – 1)
= {√7[((√3)^n – 1)] (√3 +1)}/[(√3)^2 – (1)^2]
= √7/2 [(√3)^n – 1) (√3 + 1)
(vii) Here, a = x^3, r = x^2
Using, Sn = a(r^n – 1)/(r – 1), we get,
Sn = x^3[(x^2)^n – 1]/(x^2 – 1)
= x^3[(x^2n) – 1)]/(x^2 – 1)
(viii) Here, a =7, r = -1/7, n = ∞
Using, S∞ = a/(1 – r), we get,
S∞ = 7/[1 – (-1/7)]
= 7/[1 + 1/7]
= 7/(8/7)
= 7 x 7/8
= 49/8
(ix) Here, a = √2, r = -1/2, n = ∞
Using, S∞ = a/(1 – r), we get,
S∞ = √2/[1 – (-1/2)]
= √2/(1 + 1/2)
= √2/(3/2)
= √2/1 x 2/3
= 2√2/3
Q.2 Find the sum of the series 81 – 27 + 9 -…-1/27.
Ans. Given, a = 81, r = -1/3 (r < 1), l = -1/27
Using, Sum = (a – lr)/(l – r)
Sn = [81 – (-1/27) x (-1/3)]/(1 – (-1/3)]
= (81 – 1/81)/(1 + 1/3)
= [(81^2 – 1)/81]/[4/3]
= (6561 – 1)/27 x 1/4
= 6560/(27 x 4)
= 1640/27
Q.3 Write the first four terms of a geometric series for which S8 = 39360 and r = 3.
Ans. Here, S8 = 39360, r = 3, n = 8
Using, Sn = a(r^n – 1)/(r – 1), we get,
S8 = a(3^8 – 1)/(3 – 1)
39360 = a(6561 – 1)/2
39360 x 2 = a(6560)
(39360 x 2)/6560 = a
6 x 2 = a
12 = a
A1 = 12
A2 = 12 x 3 = 36
A3 = 36 x 3 = 108
A4 = 108 x 3 = 324
Q.4 How many terms of the G.P. 32, 16, 8, … are needed to give the sum 63 3/4?
Ans. Here, a = 32, r = 1/2 (r<1), Sn = 63 3/4 = 255/4
using, Sn = a(1 – r^n)/(1 – r), we get,
255/4 = 32[1 – (1/2)^n]/(1 – 1/2)
255/4 = 32[1 – 1/2^n]/(1/2)
255/(4×32) = [1 – 1/2^n]^(1/2) x 2/1
255/(2^2×2^5×2^1) = (2^n – 1)/2^n
255/2^8 = (2^n – 1)/2^n
(256 – 1)/2^8 = (2^n – 1)/2^n
(2^8 – 1)/2^8 = (2^n – 1)/2^n
n = 8
Q.5 The sum of the first two terms of a G.P. is 36 and the product of first term and third term is 9 times the second term. Find the sum of first 8 terms.
Ans. S2 = 36, a1 x 13 = 9a2
a x ar^2 = 9 x ar
ar = 9
a = 9/r
Now, a(r^2 – 1)/(r – 1) = 36
9/r(r^2 – 1)/(r – 1) = 36
[(r – 1)(r + 1)]/[r(r – 1)] = 4
r + 1 = 4r
1 = 3r
r = 1/3
Therefore, a = 9/(1/3)
= 27
In the G.P, a = 27, r = 1/3, n = 8, Sn = ?
Using, Sn = a(1 – r^2)/(1 – r), we get,
S8 = 27[1 – (1/3)^8]/[1 – 1/3]
= 3^3[1 – (1/3)^8]/[2/3]
= 3^3[(3^8 – 1)]/3^8 x 3/2
= [(6561 – 1)]/3^5 x 3/2
= 6560/81 x 1/2
= 3280/81
Q.6 The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.
Ans. L = an = 128, Sn = 225, r = 2
Sn = (a – Lr)/(1 – r)
255 = (a x 128 x 2)/(1 – 2)
-255 = a – 256
a = 1
Q.7 If the sum of first six terms of a G.P. is equal to 9 times the sum of the first three terms, then find the common ratio of the G.P.
Ans. S6 = 9 x S3
Using, Sn = a(r^n – 1)/(r – 1), we get,
a(r^6 – 1)/(r – 1) = 9a(r^3 – 1)/(r – 1)
(r^6 – 1) = 9(r^3 – 1)
r^6 – 1 = 9r^3 – 9
r^6 – 9r^3 – 1 + 9 = 0
r^6 – 9r^3 + 8 = 0
[r^3]^2 – 9r^3 + 8 = 0
Let, r^3 = x
x^2 – 9x + 8 = 0
x^2 – 8x – 1x + 8 = 0
x(x – 8) – 1(x – 8) = 0
(x – 8) (x – 1) = 0
x = 8 or 1
Now, x = r^3 = 8
Therefore, r = 2
Q.8 How many terms of the sequence 3, 3^2, 3^3, … are needed to give the sum 120?
Ans. Here, a = 3, r = 3, Sn = 120, n = ?
Using, Sn = a(r^n – 1)/(r – 1), we get,
120 = 3(3^n – 1)/(3 – 1)
120/3 = (3^n – 1)/2
40 x 2 = 3^n – 1
80 + 1 = 3^n
81 = 3^n
3^4 = 3^n
Therefore, n = 4
Q.9 (i) If the second term of a G.P. is 2 and the sum to infinity is 8, find the first term.
(ii) The first term of an infinite G.P. is 1 and any term is equal to the sum of all the succeeding terms. Find the G.P.
Ans. (i) Here, a2 = 2, ar = 2, r = 2/a, S∞ = 8
Using, S∞ = a/(1 – r)
8 = a/(1 – r)
8 = a/(1 – 2/a)
8 = [a/(a-2)/2]
8 = a^2/(a – 2)
a^2 = 8a – 16
a^2 – 8a + 16 = 0
(a – 4)^2 = 0
a – 4 = 0
a = 4
(ii) Here, a = 1, an = a(n+1) + a(n+1) + a(n+3)…….∞.
an = a(n+1)/(1 – r) [S∞ = a/(1 – r)]
ar^(n-1) = a x r^n/(1 – r)
r^n/r^1 = r^n/(1 – r)
1 – r = n
1 = r + n
1 = 2r
1/2 = r
therefore, G.P : 1, 1/2, 1/4, 1/8……
Q.10 Evaluate 3 x 3^1/2 x 3^1/4 x 3^1/8 x … to ∞.
Ans. 3^[1+1/2+1/4+1/8…..∞
Power of 3 for an infinite G.P
Using, S∞ = a/(1 – r), we get,
3^[S∞] here, in the G.P
a = 1, r = 1/2
= 3^[1/(1-1/2)]
= 3^[1/(1/2)]
= 3^[1×2/1]
= 3^2
= 9
11. Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728.
Ans. Here, r = 3, L = 486, Sn = 728, a = ?
Using, Sn = (Lr – a)/(r – 1), we get,
728 = (486 x 3 – a)/(3 – 1)
728 = (486 x 3 – a)/2
728 x 2 = 486 x 3 – a
1456 = 1458 – a
a = 1458 – 1456
= 2
Q.12 If the first term of a G.P. is 5 and the sum of first three terms is 31/5 find the common ratio.
Ans. Here, a = 5, S3 = 31/5
Using, Sn = a(r^n – 1)/(r – 1), we get,
31/5 = 5(r^3 – 1)/(r – 1)
31/25 = (r^3 – 1^3)/(r – 1)
[Using, (a^3 – b^3) = (a – b) (a^2 + ab + b^2]
31/25 = [(r – 1)(r^2 + r + 1)/(r – 1)]
r^2 + r + 1 = 31/25
25r^2 + 25r + 25 = 31
25r^2 + 25r + 25 – 31 = 0
25r^2 + 25r – 6 = 0
25r^2 + 30r – 5r – 6 = 0
5r(5r + 6) – 1(5r + 6) = 0
(5r + 6) (5r – 1) = 0
r = -6/5 or 1/5
Q.13 The sum of first three terms of a G.P. is to the sum of first six terms as 125 : 152. Find the common ratio of the G.P.
Ans. Here, S3/S6 = 125/152
Using, Sn = a(r^n – 1)/(r – 1), we get,
[a(r^3 – 1)/(r – 1)]/[a(r^6 – 1)/r – 1)] = 125/152
(r^3 – 1)/[(r^3)^2 – (1)^2] = 125/152
(r^3 – 1)/[(r^3 – 1)(r^3 + 1)] = 125/152
152 = 125r^3 + 125
27 = 125r^3
(3)^3 = (5r)^3
3 = 5r
r = 3/5
Q.14 Evaluate: ∑^10 (k = 1) (3+2^k)
Ans. (3 + 2^1) + (3 + 2^2) +…….+ (3 + 2^10)
(3 + 3 + 3 +…….+ 3) + (2 + 2^2 +………+ 2^10)
(3 x 10) + 2(2^10 – 1)/(2 – 1) [Using, Sn = a(r^n – 1)/(r – 1)]
= 30 + 2^11 – 2
= 28 + 2^11
Q.15 Find the sum of n terms of a series whose mth term is 2m + 2m.
Ans. Here, am = 2^3 + 2m
a1 = 2^1 + 2 = 2
a2 = 2^2 + 4 = 8
a3 = 2^3 + 6 = 14
a4 = 2^4 + 8 = 24
Here, (2^1 + 2) + (2^2 + 2 x 2) + (2^3 + 2 x 3) + (2^4 + 2 x 4) + …
[2^1 + 2^2 + 2^3 + 2^4 +…..] + [2 + 4 + 6 + 8+…..]
Using, Sn = a(r^n – 1)/(r – 1), we get,
2(2^n – 1)/(2 – 1) + 2[1 + 2 + 3 + 4 +….n]
2(2^n – 1) + 2[n(n+1)]/2
2(2^n – 1) + n(n + 1)
Q.16 Find the sum of the following series to n terms:
(i) 7 + 77 + 777 +…
(iii) 0.5 + 0.55 + 0.555 +…
(ii) 8 + 88 + 888 +….
Ans. (i) Sn = 7 + 77 + 777 +…….n terms
= 7(1 + 11 + 111 +…….n terms)
= 7/9(9 + 99 + 999 +……..n terms)
= 7/9[(10 – 1) + (10^2 – 1) + (10^3 – 1) +….n terms)
= 7/9[(10 + 10^2 + 10^3 +…. n terms) – (1 + 1 + 1 +….n terms)
= 7/9[10(10^n – 1)/(10 – 1) – n]
= 7/9[(10^n+1) – 10)/9 – n]
= 7/9[(10^n+1 – 10 – 9n)/9]
= 7/81 (10^n+1 – 10 – 9n)
(ii) 8 (1 + 11 +111 +….)
8/9 (9 + 99 +999 +……)
8/9 [(10-1) + (100-1) + (1000-1) +…..]
8/9 [(10 + 10^2 + 10^3 + ….) – (1 + 1 + 1 +…..)]
8/9 [10(10^n -1)/(10 – 1) – n]
8/9 [(10^n+1 – 10 – n)/9]
8/9 [(10^n – 10 – 9n)/9]
8/81 (10^n+1 – 9n – 10)
(iii) 5 [0.1 + 0.11 + 0.111 +…..]
5/9 [0.9 + 0.99 + 0.999 +……]
5/9[(1 – 0.1) + (1 – 0.01) + )1 – 0.001) +…..]
5/9 [(1 + 1 + 1 +…) – (0.1 + 0.01 + 0.001 +…..]
5/9 [n – (1/10 – 1/10^2 – 1/10^3 +…..]
Clearly, it forms a G.P
a = 1/10, r = 1/10
5/9 [n – (1/10[1 – (1/10)^n])/(1 – (1/10))
5/9 [n – (1 – 1/10^n)/10 x 10/9]
5/9 [n – (1 – 1/10^n)/9]
5/9 [(9n – 1 + 1/10^n)/9]
5/81 [9n – 1 + 1/10^n]
Q.17 Using geometric series write 0.175175175175175… in fraction.
Ans. x = 0.175 + 0.000175 + 0.000000175 +……..
x = 175/10^3 + 175/10^6 + 175/10^9 +……
x = 175 [1/10^3 + 1/10^6 + 1/10^9 +……]
Clearly, they form a G.P
x = 175 [(1/10^3)/(1 – 1/10^3)]
x = 175 [(1/10^3)/((10^3-1)/10^3]
x = 175 [1/10^3 x 10^3/(1000 -1)]
x = 175/999
Q.18 The sum of an infinite G.P. is 16 and the sums of the squares of its terms is 153 3/5. Find the common ratio and fourth term of the progression.
Ans. Let, G.P be a, ar, ar^2, ar^3……
Here, S∞ = 16
a/(1-r) = 16
a = 16(1 – r)
Now,
a^2 + a^2r^2 + a^2r^4-1 +…….∞ = 153 3/5
a^2/(1 – r^2) = 768/5
[16^2(1 – r)^2]/[(1 – r)(1 + r)] = 768/5
256 (1 -r)/(1 + r) = 768/5
(1 – r)/(1 + r) = 3/5
5 – 5r = 3 + 3r
-8r = -2
r = 1/4
Therefore, a = 16(1 – 1/4)
= 16 x 3/4
= 12
a4 = ar^3
= 12 x 1/64
= 3/16
Q.19 A snail starts moving towards a point 3 cm away at a pace of 1 cm per hour. As it gets tired, it covers only half the distance compared to previous hour in each succeeding hour. In how much time will the snail reach his target?
Ans. Distance = 1, 1/2, 1/4, 1/8…….
Clearly, distance covered by snail forms a G.P
a = 1, r = 1/2
Using, S∞ = a/(1 – r)
= 1/(1 – 1/2)
= 1/(1/2)
= 2 cm
Q.20 The inventor of chessboard was a very clever man. He asked the king, a reward of one grain of wheat for the first square, 2 grains for the second, 4 grains for the third, and so on doubling the amount of the grain for each subsequent square. How many grains would have to be given?
Ans. Here, Reward for sq. 1 = 1 grain
Reward for sq. 2 = 2 g
Reward for sq. 3 = 4 g
Reward for sq. 4 = 8 g
Clearly, the pattern of rewards form a G.P
Total reward = 1 + 2 + 4 + 8 +….
Here, a = 1, r = 2, n = 64
Using, Sn = a(r^n – 1)/(r – 1), we get,
S64 = 1(2^64 – 1)/(2 – 1)
= 2^64 – 1
Q.21 On the first day, a music video of Arjit Singh was posted online, got 120 views in Delhi. The number of viewership gets increased by 5% per day. How many total views did the video get over the course of the first 29 days? Express your answer in exponential form.
Ans. a = 120, r = 105% = 105/100 = 1.05, L = 120(1.05)^28, n = 29
Sn = (Lr – a)/(n – 1)
S29 = [120(1.05)^28 x (1.05) – 120]/[1.05 – 1]
= [120(1.05)^29 – 120]/0.05
= 120[(1.05)^29 – 1] x 100/5
= 2400 [(1.05)^29 – 1]
Q.22 A person sends a fake news on WhatsApp to 4 of his friends on Monday. Each of these friends forward the fake news onto the four of their friends on Tuesday. Each person who receives the fake news on Tuesday send it to four more people by Wednesday and this process goes on for a week. After a week a leading news agency verifies the news and finds it to be fake. Find how many people have received the fake news on WhatsApp till then?
Ans. Monday = 4, Tuesday = 16, Wednesday = 64
Continuous for 7 days.
Clearly, they form a G.P, with,
a = 4, r = 4, n = 7
Total recipient = 4 + 16 + 64 + ………
TR – S7
Using, Sn = a(r^n – 1)/(r – 1), we get,
S7 = 4(4^7 – 1)/(4 – 1)
= 4/3 [16384 – 1]
= 4/3 x 16383
= 21844
Q.23 You and your sibling decide to ask for a raise in your pocket money from your dad. Your dad gives both of you a chance. You two could have Rs.1000 at once or can get Rs.2 on day one, Rs.4 on day two and so on receiving twice as many rupees each day as the previous day for 12 days. While you opted for getting Rs.1000 at once and your brother opted for the later. Which one of you made a better decision and why?
Ans. Here,
Raise on day 1 = 2
Raise on day 2 = 4
Raise on day 3 = 8
Raise for 12 days….
Clearly, the raise form a G.P.
With, a = 2, r = 2, n = 12
Total raise = 2 + 4 + 8 +…..for 12 days.
Using, Sn = a(r^n – 1)/(r – 1)
S12 = 2(2^12 – 1)/(2 – 1)
= 2(4096 – 1)
= 2 x 4095
= Rs.8190
Q.24 Due to reduced taxes, an individual has an extra Rs.30000 is spendable income. If we assume that an individual spends 70% of this on consumer goods and the producers of these goods in turn spends 70% on consumer goods and this process continues indefinitely. What is the total amount spent on consumer goods?
Ans. a1 = 30000 x 70/100 = 21000
a2 = 21000 x 70/100 = 21000 x 0.7
a3 = 21000 x 0.7 x 70/100 = 21000 x (0.7)^2
S∞ = a/(1 – r)
= 21000/(1 – 0.7)
= 21000/0.3
= 210000/3
= 70000
Q.25 A pendulum swings through an arc of 25 cm. On each successive swing, the pendulum covers an arc equal to 90% of the previous swing. Find the length of the arc on the sixth swing and the total distance the pendulum travels before coming to rest.
Ans. a = 25, a2 = 25 (0.9), a3 = 25 (0.9)^2
Using, an = ar^(n-1), we get,
a6 = 25 x (0.9)^5
= 25 x 0.59049
= 14.76225
= 14.76 cm
Now, pendulum coming to rest would form a infinite G.P.
Using, S∞ = a/(1 – r)
= 25/(1 – 0.9)
= 25/0.1
= 250 cm
Q.26 A square is drawn by joining the mid-points of the sides of a square. A third square is drawn inside the second square by joining the mid-points of the second square and the process is continued indefinitely. If the side of the original square is 8 cm, find the sum of the areas of all the squares.
Ans. In ΔAPB, we have ∠ = 90° (avg of square)
Using, PT, we get,
AP^2 + PB^2 = AB^2
4^2 + 4^2 = AB^2 [AB = 4√2 cm]
16 + 16 = AB^2
32 = AB^2
√32 = AB
√16×2 = AB
Area of sq. 1 = 8^2 = 64cm^2 (A1)
Area of sq. 2 = (4√2)^2 cm = 32cm^2 (A2)
Area of sq. 3 = (4^2) = 16cm^2 (A3)
Clearly, they form a G.P with a = 64, r = 1/2
Sum of areas = S∞ (infinite G.P)
Using, S∞ = a/(1 – r), we get,
= 64/(1 – 1/2)
= 64/(1/2)
= 64 x 2/1
= 128 cm^2
Q.27 A ball is dropped from a height of 90 feet and always rebounds one-third the distance from which it falls. Find the total vertical distance, the ball travelled when it hits the ground for the third time. Also, find the total distance travelled by the ball before it comes to rest.
Ans. Distance covered at first bounce = 90 feet
Distance covered at second bounce = 90 x 1/3
= 30 feet x 2 = 60 feet
Distance covered at third bounce = 30 x 1/3 = 10 feet x 2
= 20 feet
Distance covered by ball when is hits the ground third time = 90 + 60 + 20 = 170 feet
Total distance = 90 + S∞ (with a =60, r = 1/3)
= 90 + 60/(1 – 1/3)
= 90 + 60/(2/3)
= 90 + 60 x 3/2
= 90 + 90
= 18 feet
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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