Inter 1st Year Maths Sequences and Series Solutions Exercise 8c

Practicing the AP Board Solutions Class 11 Maths and Chapter 8 Inter 1st Year Maths Sequences and Series Solutions Exercise 8c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Sequences and Series Solutions Exercise 8c

Sequences and Series Exercise 8c Solutions

Sequences and Series Class 11 Exercise 8c Solutions – Sequences and Series 8c Exercise Solutions

I.

Question 1.
Find the 20th term of the series 2x⁴ + 4x⁶ + 6x⁸ + … + n terms.
Solution:
Given series is 2x⁴ + 4x⁶ + 6x⁸ + … + n terms
nth term = a_n = 2n × (2n + 2) = 4n² + 4n
The 20th term, a₂₀ = 4(20)² + 4(20)
= 4(400) + 80
= 1600 + 80
= 1680
Therefore, the 20th term of the series is 1680.

Question 2.
The sum of some terms of G.P. is 315, whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Solution:
Given that the sum of some terms in a G.P. is 315.
Let the number of terms be n.
We know that the sum of terms is
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\)
Given that the first term a is 5 and the common ratio r is 2.
315 = \(\frac{5\left(2^n-1\right)}{2-1}\)
⇒ 2ⁿ – 1 = 63
⇒ 2ⁿ = 64 = (2)⁶
⇒ n = 6
Hence, the last term of the G.P. = 6th term
= ar^6-1
= (5) (2)⁵
= (5) (32)
= 160
Therefore, the last term of the G.P. is 160.

Question 3.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Solution:
Let’s consider a and r to be the first term and the common ratio of the G.P., respectively.
Given, a = 1
a₃ = ar² = r²
a₅ = ar⁴ = r⁴
Then, from the question we have
r² + r⁴ = 90
r⁴ + r² – 90 = 0
r² = \(\frac{-1 \pm \sqrt{1+360}}{2}\)
= \(\frac{-1 \pm \sqrt{361}}{2}\)
= \(\frac{-1 \pm 19}{2}\)
= -10 or 9
∴ r = ±3
Therefore, the common ratio of the G. P. is ±3.

II.

Question 1.
The sum of three consecutive terms in a G.P. is 56. If we subtract 1, 7, and 21 from those numbers in that order, we obtain an arithmetic progression. Find the numbers.
Solution:
Let’s consider the three numbers in G.P. to be a, ar, and ar².
Then, from the question, we have
a + ar + ar² = 56
a(1 + r + r²) = 56
Also given a – 1, ar – 7, ar² – 21 from an A.P.
So, (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)
⇒ ar – a – 6 = ar² – ar – 14
⇒ ar² – 2ar + a = 8
⇒ ar² – ar – ar + a = 8
⇒ a(r² + 1 – 2r) = 8
⇒ a(r – 1)² = 8
⇒ \(\frac{56}{1+r+r^2}\) (r – 1)² = 8
⇒ 7(r² – 2r + 1) = 1 + r + r²
⇒ 7r² – 14r + 7 – 1 – r – r² = 0
⇒ 6r² – 15r + 6 = 0
⇒ 6r² – 12r – 3r + 6 = 0
⇒ 6r(r – 2) – 3(r – 2) = 0
⇒ (6r – 3) (r – 2) = 0
⇒ r = 2, \(\frac {1}{2}\)
when r = 2, a = 8
when r = \(\frac {1}{2}\), a = 32
Thus, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When r = \(\frac {1}{2}\), the three numbers in G.P. are 32, 16, and 8.
Therefore, in either case, the required three numbers are 8, 16, and 32.

Question 2.
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Solution:
Let’s consider the terms in the G.P. to be T₁, T₂, T₃, T₄, and so on. T_2n.
The number of terms = 2n
Then, from the question we have
T₁ + T₂ + T₃ + ….. + T_2n = 5[T₁ + T₃ + ….. + T_2n-1]
⇒ T₁ + T₂ + T₃ + ….. + T_2n – 5[T₁ + T₃ + ….. + T_2n-1] = 0
⇒ T₂ + T₄ + …… + T_2n = 4[T₁ + T₃ + ….. + T_2n-1] ……….(1)
Now, let the terms in G.P. be a, ar, ar², ar³, ………
Then (1) becomes,
\(\frac{{ar}\left(r^n-1\right)}{r-1}=\frac{4 \times a\left(r^n-1\right)}{r-1}\)
⇒ ar = 4a
⇒ r = 4
Thus, the common ratio of the G.P. is 4.

Question 3.
If \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\), (x ≠ 0) then show that a, b, c, and d are in G.P.
Solution:
Given \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}\)
On cross-multiplying, we have
(a + bx) (b – cx) = (b + cx) (a – bx)
⇒ ab – acx + b²x – bcx² = ab – b²x + acx – bcx²
⇒ 2b²x = 2acx
⇒ b² = ac
⇒ \(\frac{b}{a}=\frac{c}{b}\) ………(1)
\(\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\)
Also, given that on cross-multiplying, we have
(b + cx)(c – dx) = (b – cx)(c + dx)
⇒ bc – bdx + c²x – cdx² = bc + bdx – c²x – cdx²
⇒ 2c²x = 2bdx
⇒ c² = bd
\(\frac{c}{d}=\frac{d}{c}\) ……….(2)
From (1) and (2), we get
\(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\)
Therefore, a, b, and c are in G.P.

Question 4.
Let S be the sum, P the product, and R the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ.
Solution:
Let the terms in G.P. be a, ar², ar³,……..ar^n-1,……..
From the question, we have

Hence P²Rⁿ = Sⁿ

Question 5.
If a, b, c, d are in G.P., prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.
Solution:
Given that a, b, c, and d are in G.P.
So, we have
b² = ac ………(i)
c² = bd ………(ii)
ad = bc ………(iii)
Required to prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are G.P.
i.e., (bⁿ + cⁿ) = (aⁿ + bⁿ) (cⁿ + dⁿ)
Taking L.H.S.

Therefore, (aⁿ + bⁿ), (bⁿ + cⁿ), and (cⁿ + dⁿ) are in G.P.
Hence proved.

Question 6.
If a and b are the roots of x² – 3x + p = 0 and c and d are the roots of x² – 12x + q = 0, where a, b, c, d form a G.P., prove that (q + p) : (q – p) = 17 : 15.
Solution:
Given that a and b are the roots of x² – 3x + p = 0
So, we have a + b = 3 and ab = p …….(i)
Also, c and d are the roots of x² – 12x + q = 0
So, c + d = 12 and cd = q ……….(ii)
And given that a, b, c, and d are in G.P.
Let’s take a = x, b = xr, c = xr², d = xr³
From (i) and (ii), we get
x + xr = 3
x(1 + r) = 3
and, xr² + xr³ = 12
xr² (1 + r) = 12
On dividing, we get
\(\frac{\mathrm{xr}^2(1+\mathrm{r})}{\mathrm{x}(1+\mathrm{r})}=\frac{12}{3}\)
⇒ r² = 4
⇒ r = ±2
when r = 2, x = \(\frac{3}{(1+2)}=\frac{3}{3}\) = 1
when r = -2, x = \(\frac{3}{(1-2)}=\frac{3}{-1}\) = -3
Case I: When r = 2 and x = 1,
ab = x²r = 2
cd = x²r⁵ = 32
\(\frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}\)
(q + p) : (q – p) = 17 : 15
Case II: When r = -2 and x = -3,
ab = x²r = -18
cd = x²r⁵ = -288
\(\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}\)
(q + p) : (q – p) = 17 : 15
Therefore, in both the cases, we get (q + p) : (q – p) = 17 : 15

Question 7.
The ratio of the A.M. and G.M. of two positive numbers a and b is m : n. Show that a : b = \(\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)\)
Solution:
Let the two numbers be a and b.
A.M. = \(\frac{(a+b)}{2}\) and G.M. = √ab
From the question, we have
\(\frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}\)
⇒ \(\frac{(a+b)^2}{4(a b)}=\frac{m^2}{n^2}\)

III.

Question 1.
Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + ……
(ii) .6 + .66 + .666 + ……..
Solution:
(i) Given 5 + 55 + 555 + ……
Let S_n = 5 + 55 + 555 + ….. + up to n terms

(ii) Given .6 + .66 + .666 + …..
Let S_n = 0.6 + 0.66 + 0.666 + ….. + up to n terms
= 6[0.1 + 0.11 + 0.111 + ….. to n terms]
= \(\frac {6}{9}\) [0.9 + 0.99 + 0.999 + ……. to n terms]

Question 2.
If f is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and \(\sum_{x=1}^n f(x)\) = 120, find the value of n.
Solution:
Given that
f(x + y) = f(x) × f (y) for all x, y ∈ N…. (1)
f(1) = 3
Taking x = y = 1 in (1), we have
f(1 + 1) = f(2)
= f(1) f(1)
= 3 × 3
= 9
Similarly, f(1 + 1 + 1) = f(3)
= f(1 + 2)
= f(1) f (2)
= 3 × 9 = 27
And, f(4) = f(1 + 3)
= f(1) f(3)
= 3 × 27
= 81
Thus, f(1), f(2), f(3) ……. that is 3, 9, 27,…….. form a G.P. with the first term and common ratio both equal to 3.
We know that the sum of terms in a G.P. is given by
S_n = \(\frac{a\left(r^n-1\right)}{r^{\prime}-1}\)
and it’s given that, \(\sum_{x=1}^n f(x)\) = 120
Hence, the sum of the terms of the function is 120.
120 = \(\frac{3\left(3^n-1\right)}{3-1}\)
⇒ 120 = \(\frac{3}{2}\left(3^n-1\right)\)
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 81
⇒ 3ⁿ = 3⁴
⇒ n = 4
Therefore, the value of n is 4.

Question 3.
A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual instalments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
Solution:
Given, the farmer pays Rs. 6000 in cash.
So, the unpaid amount = Rs. 12000 – Rs. 6000 = Rs. 6000
From the question, the interest paid annually will be 12% of 6000, 12% of 5500, 12% of 5000,………… 12% of 500.
Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + ……. 12% of 500
= 12% of (6000 + 5500 + 5000 + ….. + 500)
= 12% of (500 + 1000 + 1500 + …….. + 6000)

Question 4.
Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual instalments of Rs. 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
Solution:
Given, Shamshad Ali buys a scooter for Rs. 22000 and pays Rs. 4000 in cash.
So, the unpaid amount = Rs. 22000 – Rs. 4000 = Rs. 18000
From the question, it’s understood that the interest paid annually = 10% of (18000 + 17000 + 16000 + …… + 1000)
= 10% of (1000 + 2000 + 3000 + ……. + 18000)
It’s seen that 1000, 2000, 3000,……….
18000 from an A.P. with the first term and common difference both equal to 1000.
Let the number of terms be n.
So, 18000 = 1000 + (n – 1) (1000)
n = 18
Now, the sum of the A.P. is given by
1000 + 2000 + ……… + 18000
= \(\frac {18}{2}\) [2(1000) + (18 – 1)(1000)]
= 9[2000 + 17000]
= 171000
Thus, Total interest paid = 10% of (18000 + 17000 + 16000 + …… + 1000)
= 10% Rs. 171000
= Rs. 17100
Therefore, the cost of scooter = Rs. 22000 + Rs. 17100 = Rs. 39100

Question 5.
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instructions that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter, find the amount spent on the postage when the 8th set of letters is mailed.
Solution:
The number of letters mailed from a G.P:
4, 4², …., 4⁸
First term = 4
Common ratio = 4
Number of terms = 8

It is given that the cost to mail one letter is 50 paise
∴ Cost of mailling 87380 letters = Rs. 87380 × \(\frac {50}{100}\) = Rs. 43690
Thus, the amount spent when the 8th set of letters is mailed is Rs. 43690.

Question 6.
A man deposited Rs. 10000 in a bank at the rate of 3% simple interest annually. Find the amount in the 15th year since he deposited the amount, and also calculate the total amount after 20 years.
Solution:
Given, the man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually.
Hence, the interest in first year = \(\frac {5}{100}\) × Rs. 10000 = Rs. 500
10000 + 500 + 500+ …… +500 (14 times)
So, the amount in the 15th year = Rs. 10000 + 14 × Rs. 500
= Rs. 10000 + Rs. 7000
= Rs. 17000
And the amount after 20 years = Rs. 10000 + 500 + 500 + …… + 500 (20 times)
= Rs. 10000 + 20 × Rs. 500
= Rs. 10000 + Rs. 10000
= Rs. 20000
Therefore, the amount in the 15th year is Rs. 17,000, and the total amount after 20 years will be Rs. 20,000.

Question 7.
A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Solution:
Given, the cost of the machine = Rs. 15625
Also, given that the machine depreciates by 20% every year.
Hence, its value after every year is 80% of the original cost,
i.e., \(\frac {4}{5}\)th of the original cost.
I5625 × \(\frac{4}{5} \times \frac{4}{5} \times \ldots \times \frac{4}{5}\) (5 times)
Therefore, the value at the end of 5 years = 5 × 1024 = 5120.
Thus, the value of the machine at the end of 5 years will be Rs. 5120.

Question 8.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day, 4 more workers dropped out on the third day, and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Solution:
Let’s assume x to be the number of days in which 150 workers finish the work.
Then, from the question, we have
150x = 150 + 146 + 142 + …….. + (x + 8) terms
The series 150 + 146 + 142 + ……….. + (x + 8) terms is an A.P.
first term (a) = 150
common difference (d) = -4
and number of terms (n) = (x + 8)
Now, finding the sum of terms
150x = \(\frac{(x+8)}{2}\) [2(150) + (x + 8 – 1)(-4)]
⇒ 150x = (x + 8) [150 + (x + 7)(-2)]
⇒ 150x = (x + 8) [150 – 2x – 14]
⇒ 150x = (x + 8) [136 – 2x]
⇒ 75x = (x + 8) [68 – x]
⇒ 75x = 68x – x² + 544 – 8x
⇒ x² + 75x – 60x – 544 = 0
⇒ x² + 15x – 544 = 0
⇒ x² + 32x – 17x – 544 = 0
⇒ x(x + 32) – 17(x + 32) = 0
⇒ (x – 17) (x + 32) = 0
⇒ x = 17 or x = -32
As x cannot be negative [number of days is always a positive quantity]
∴ x = 17
Hence, the number of days in which the work should have been completed is 17.
But, due to the dropping out of workers, the number of days in which the work is completed = (17 + 8) = 25.