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Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Miscellaneous Exercise On Chapter 9 | Set 2
  • Last Updated : 01 Apr, 2021

Question 17.  If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.

Solution: 

We are given, a, b, c and d are in G.P.

Therefore, we have

b2 = ac … (1)

c2 = bd … (2)

ad = bc … (3)

We need to prove (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,

=> (bn+ cn)2 = (an + bn) (cn + dn)

Solving L.H.S., we get

= b2n + 2bncn + c2n

= (b2)n + 2bncn + (c2)n

= (ac)n + 2bncn + (bd) [From (1) and (2)]

= ancn + bncn+ bncn + bndn

= ancn + bncn+ andn + bndn  [From (3)]

= cn(an + bn) + dn(an + bn)

= (an + bn) (cn + dn)

= R.H.S.

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

Hence, proved.

Question 18.  If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that 

(q + p):(q – p) = 17:15.

Solution: 

We are given that a, b, c, d are in G.P. Let’s suppose the common ratio is r.

So, b=ar, c=ar2 and d=ar3

Now a and b are the roots of x2 – 3x + p = 0.

 Sum of roots = a + b = 3 

=> a + ar = 3 

=> a(1+r) = 3 ….. (1)

 Product of roots = ab = p

=> a(ar) = p

=> a2r = p  ….. (2)

And c, d are the roots of x2 − 12x + q = 0

Sum of roots = c + d = 12

=> ar2 + ar3 = 12

=> ar2(1+r) = 12  ….. (3)

Product of roots = cd = q

=> ar2(ar3) = q

=> a2r5 = q  ….. (4)

Dividing equation (3) by (1),we get,

=> \frac{ar^2(1+r)}{a(1+r)}  = \frac{12}{3}

=> r2 = 4

=> r = ±2

When r=2, from (1), we get,

=> a(1+2) = 3

=> a = 1

Putting a=1 and r=2 in (2),

=> p = (1)2(2) = 2

From (4) we get,

q = (1)2(2)5 = 32



Now L.H.S. = \frac{q+p}{q−p}  = \frac{32+2}{32−2}  = \frac{34}{30}  \frac{17}{15}  

= R.H.S. 

When r=−2, from (1), we get,

=> a(1−2) = 3

=> a = −3

Putting a=−3 and r=−2 in (2),

p = (−3)2(−2) = −18

From (4) we get,

q = (−3)2(−2)5 = −288

Now L.H.S. = \frac{q+p}{q−p}  = \frac{−288+(−18)}{−288−(−18)}  = \frac{−306}{−270}  = \frac{17}{15}

= R.H.S.

Hence, proved.

Question 19. The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a:b = (m+√(m2−n2)):(m-√(m2-n2)).

Solution:

We are given two numbers a and b. Therefore,

A.M = (a + b)/2 and G.M. = \sqrt{(ab)}

It’s given that, \frac{a+b}{2\sqrt{(ab)}}  = \frac{m}{n}

Applying Componendo and Dividendo on both sides, we get,

=> \frac{a+b+2\sqrt{(ab)}}{a+b−2\sqrt{(ab)}}  = \frac{m+n}{m−n}

=> \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}−\sqrt{b})^2}  = \frac{m+n}{m−n}

=> \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}−\sqrt{b}} = \frac{\sqrt{(m+n)}}{\sqrt{(m−n)}}

By again applying Componendo and Dividendo on both sides, we get,

=> \frac{(\sqrt{a}+\sqrt{b}+\sqrt{a}−\sqrt{b})}{(\sqrt{a}+\sqrt{b}−\sqrt{a}+\sqrt{b})}  = \frac{\sqrt{(m+n)}+\sqrt{(m−n)}}{\sqrt{(m+n)}−\sqrt{(m−n)}}



=> \frac{\sqrt{a}}{\sqrt{b}}  = \frac{\sqrt{(m+n)}+\sqrt{(m−n)}}{\sqrt{(m+n)}−\sqrt{(m−n)}}

Squaring both sides, we get,

=>\frac{a}{b}  = \frac{(m+n)+(m-n)+2\sqrt{(m+n)}\sqrt{(m−n)}}{(m+n)+(m-n)-2\sqrt{(m+n)}\sqrt{(m−n)}}

=> \frac{a}{b}  = \frac{2m+2\sqrt{(m^2−n^2)}}{2m-2\sqrt{(m^2−n^2)}}

=> \frac{a}{b}  = \frac{m+\sqrt{(m^2−n^2)}}{m-\sqrt{(m^2−n^2)}}

Hence, proved.

Question 20. If a, b, c are in A.P,; b, c, d are in G.P and 1/c, 1/d, 1/e are in A.P. prove that a, c, e are in G.P.

Solution:

We are given that a, b, c are in A.P.

Hence, b – a = c – b  ….. (1)

And, given that b, c, d are in G.P.

So, c2 = bd  ….. (2)

Also, we know 1/c, 1/d, 1/e are in A.P. Therefore,

=> 1/d – 1/c = 1/e – 1/d

=> 2/d = 1/c + 1/e   ….. (3)

We need to prove that a, c, e are in G.P.

From (1), we get

=> 2b = a + c

=> b = (a + c)/ 2

And from (2), we get

=> d = \frac{c^2}{b}

On putting above values in (3), we get,

=> \frac{2b}{c^2}  = 1/c + 1/e

=> \frac{2(a+c)}{2c^2}  = 1/c + 1/e

=> \frac{a+c}{c^2}  \frac{c+e}{ce}

=> \frac{a+c}{c}  = \frac{c+e}{e}

=> (a+c)e = (c+e)c

=> ae+ce = c2+ec

=> c2 = ae

Therefore, a, c, e are in G.P.

Question 21. Find the sum of the following series up to n terms:

(i) 5 + 55 + 555 + …

Solution:

Let us take Sn = 5 + 55 + 555 + ….. up to n terms

Multiplying and dividing by 9, we get

\frac{5}{9} [9+99+999+…to n terms]

\frac{5}{9} [(10–1)+(102–1)+(103–1)…to n terms]

\frac{5}{9} [(10+102+103+….n terms) – (1+1+1+…..n terms)]

\frac{5}{9}
\left[\frac{10(10^n-1)}{10-1}-n \right]

\frac{5}{9}
\left[\frac{10(10^n-1)}{9}-n\right]

\frac{50(10^n-1)}{81} – \frac{5n}{9}

(ii) .6 + .66 + .666 + …

Solution:

Let us take Sn = 0.6 + 0.66 + 0.666 + … up to n terms

= 6 [0.1+0.11+0.111+…up to n terms]

Multiplying and dividing by 9, we get

\frac{6}{9}  [0.9+0.99+0.999+…up to n terms]

\frac{6}{9}
\left[\left(1-\frac{1}{10} \right)+\left(1-\frac{1}{10^2}\right)+\left(1-\frac{1}{10^3}\right)+...n\hspace{0.1cm}terms\right]

\frac{2}{3}
\left[(1+1+1+...n\hspace{0.1cm}terms) - \frac{1}{10}\left(1+\frac{1}{10}+\frac{1}{10^2} +...n\hspace{0.1cm}terms\right)\right]

\frac{2}{3}
\left[n-\frac{1}{10}\left[\frac{1-\left(\frac{1}{10}\right)^n}{1-\frac{1}{10}}\right]\right]

= \frac{2}{3}  \frac{2(1-10^{-n})}{27}

Question 22. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Solution:

We are given the series: 2 × 4 + 4 × 6 + 6 × 8 + … n terms

nth term = an = 2n × (2n + 2) = 4n2 + 4n

Putting n=20, we would get the 20th term,

a20 = 4(20)2 + 4(20) = 1600 + 80 = 1680

Therefore, the 20th term of the series is 1680.

Question 23. Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

Solution:

We are given the series: 3 + 7 + 13 + 21 + 31 + …

Let’s take S as the sum of this series. Therefore,

S = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an     ….. (1)

S = 3 + 7 + 13 + 21 + …. + an–2 + an–1 + an ….. (2)

On subtracting (2) from (1), we get

=> S – S = [3 + (7 + 13 + 21 + 31 + …+ an–1 + an)] – [(3 + 7 + 13 + 21 + 31 + …+ an–1) + an]

=> 0 = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – an

=> an = 3 + [4 + 6 + 8 + … (n–1) terms]

As these series 4 + 6 + 8 + … (n–1) terms is an A.P., we can easily find its sum. Therefore,

= 3+\frac{n-1}{2}[(2×4)+(n-1-1)2]     

= 3+\frac{n-1}{2}[8+(n-2)2]

= 3+(n1)(n+2) 

= n2+n+1 

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{k=n} a_k = \sum_{k=1}^{k=n} k^2 + \sum_{k=1}^{k=n} k + \sum_{k=1}^{k=n} 1

Sn \frac{n(n+1)(2n+1)}{6}  + \frac{n(n+1)}{2}  + n

\frac{n[(n+1)(2n+1)+3(n+1)+6]}{6}

\frac{n[(n+1)(2n+1)+3(n+1)+6]}{6}

\frac{n[2n^2+6n+10]}{6}

\frac{n(n^2+3n+5)}{3}

Therefore, sum of the series is \frac{n(n^2+3n+5)}{3}.

Question 24. If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22 = S3 (1 + 8S1).

Solution:

According to the question, we have,

S1 \frac{n(n+1)}{2}

S2 \frac{n(n+1)(2n+1)}{6}

S3 \frac{n^2(n+1)^2}{4}

Therefore, L.H.S. = 9S22

= 9 \left[\frac{n(n+1)(2n+1)}{6}\right]^2

\frac{[n(n+1)(2n+1)]^2}{4}

And R.H.S. = S3(1 + 8S1)

\frac{n^2(n+1)^2}{4}  \left[1 + \frac{8n(n+1)}{2}\right]

\frac{n^2(n+1)^2}{4}  (1+ 4n2 + 4n)

\frac{[n(n+1)(2n+1)]^2}{4}

= L.H.S.

Hence, proved.

Question 25. Find the sum of the following series up to n terms: \frac{1^3}{1} + \frac{1^3+2^3}{1+3} + \frac{1^3+2^3+3^3}{1+3+5} + …

Solution:

The nth term of this series can be written as:

=> an \frac{\left[\frac{n(n+1)}{2}\right]^2}{[1+3+5+...+(2n-1)]}

Here the denominator 1+3+5+…+(2n-1) is an A.P. 

with first term(a) 1,common difference 2 and last term 2n-1.

Total number of terms will be n. So,

1+3+5+7…+(2n1) = \left(\frac{n}{2}\right)[2×1+(n-1)2] = n2

Thus, our nth term becomes,

an \frac{n^2(n+1)^2}{4n^2}

\frac{(n+1)^2}{4}

\frac{n^2}{4} + \frac{n}{2} + \frac{1}{4}

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{k=n} a_k = \sum_{k=1}^{k=n} \left(\frac{k^2}{4}+\frac{k}{2}+\frac{1}{4}\right)

 \left(\frac{1}{4}\right)
\left[\frac{n(n+1)(2n+1)}{6}\right] \left(\frac{1}{2}\right)
\left[\frac{n(n+1)}{2}\right] + \frac{n}{4}

\frac{n[(n+1)(2n+1)+6(n+1)+6]}{24}

\frac{n(2n^2+3n+1+6n+6+6)}{24}

 \frac{n(2n^2+9n+13)}{24}

Therefore, sum of the series is  \frac{n(2n^2+9n+13)}{24}.

Question 26. Show that \frac{1×2^2+2×3^2+...+n×(n+1)^2}{1^2×2+2^2×3+...+n^2×(n+1)} = \frac{3n+5}{3n+1}.

Solution:

We can see that,

The nth term of the numerator = n(n + 1)2 = n3 + 2n2 + n

Now we have to summate this nth term to find the sum(S1) of numerator. 

S_1 = \sum_{k=1}^{k=n} (k^3+2k^2+k)      

\frac{n^2(n+1)^2}{4} + \frac{2n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}

\frac{n(n+1)}{2} \left[\frac{n(n+1)}{2} + \frac{2(2n+1)}{3} + 1\right]

 \frac{n(n+1)}{2}\left[\frac{(3n^2+3n+8n+4+6)}{6}\right]

\frac{n(n+1)}{2}\left[\frac{3n^2+11n+10}{6}\right]

\frac{n(n+1)}{2}\left[\frac{(3n+5)(n+2)}{6}\right]

\frac{n(n+1)(3n+5)(n+2)}{12}

And nth term of the denominator = n2(n + 1) = n3 + n2

Now we have to summate this nth term to find the sum(S2) of denominator. 

S_2 = \sum_{k=1}^{k=n} (k^3+k^2)

 \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}

 \frac{n(n+1)}{2}\left[\frac{n(n+1)}{2} + \frac{2n+1}{3}\right]

\frac{n(n+1)}{2}\left[\frac{3n^2+3n+4n+2}{6}\right]

\frac{n(n+1)}{2}\left[\frac{3n^2+7n+2}{6}\right]

\frac{n(n+1)}{2}\left[\frac{(3n+1)(n+2)}{6}\right]

\frac{n(n+1)(3n+1)(n+2)}{12}

Now sum of the series(S) is equal to \frac{S_1}{S_2}

S = \frac{\frac{n(n+1)(3n+5)(n+2)}{12}}{\frac{n(n+1)(3n+1)(n+2)}{12}}

\frac{3n+5}{3n+1}

Hence, proved.

Question 27. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him?

Solution:

We are given that the farmer pays Rs 6000 in cash.

So, the remaining unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

According to the question, the total interest to be paid is,

Interest = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)

Here the series 6000 + 5500 + 5000 + … + 500 is an A.P. with the first term a=6000 and common difference d=–500.

Let us take the number of terms of A.P. be n.

We know nth term of an A.P. is given by, an = a+(n1)d

=> 500 = 6000+(n-1)(–500)

=> 5500 = 500n–500

=> n = 12

Now,

The sum of the A.P = \frac{12}{2}  [2(6000) + (12–1)(-500)] 

= 6 [12005500] = 39000

Thus, the total interest to be paid = 12% of (6000 + 5500 + 5000 + … + 500)

= 12% of 39000 = Rs 4680

Therefore, the tractor will cost the farmer = (Rs 12000 + Rs 4680) = Rs 16680

Question 28. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in the annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Solution:

We are given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

So, the remaining unpaid amount = Rs 22000 – Rs 4000 = Rs 18000

According to the question, the total interest to be paid,

Interest = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000

= 10% of (18000 + 17000 + 16000 + … + 1000)

Here, 18000, 17000, 16000 … 1000 forms an A.P. with first term a=18000 and common difference d=–1000.

Let’s take number of terms be n.

So, 1000 = 18000 + (n – 1) (–1000)

=> 17000 = 1000n–1000

=> n = 18

Now, the sum of the A.P. = \frac{18}{2}  [2(18000) + (18–1)(-1000)]

= 9 [36000–17000] = 9 [19000] = 171000

Thus, the total interest to be paid = 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of 171000 = Rs 17100

Thus, the cost of scooter = (Rs 22000 + Rs 17100) = Rs 39100

Question 29. 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 instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paisa to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

Solution:

The numbers of letters mailed forms a G.P. : 4, 42, … 48 where first term, a = 4 and common ratio, r = 4

And the number of terms, n = 8

The sum of n terms of a G.P. is given by, Sn = \frac{a(r^n-1)}{r-1}

\frac{4(4^8-1)}{4-1}

\left(\frac{4}{3}\right) (65536–1) = 4(21845) = 87380

Now it’s given that the cost to mail one letter is 50 paisa.

Thus, Cost of mailing 87380 letters = Rs 87380 x (50/100) = Rs 43690

Therefore, the amount spent on the postage when 8th set of letter is mailed will be Rs 43690.

Question 30. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

Solution:

Given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

Thus, the interest in first year = (5/100) x Rs 10000 = Rs 500

Now, Amount in 15th year = 10000 + (500+500+500+…14 times)

= Rs 10000 + 14 × Rs 500

= Rs 10000 + Rs 7000 = Rs 17000

Amount in 20th year = 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 17000 and the total amount after 20 years will be Rs 20000.

Question 31. 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:

We are given the cost of machine is Rs 15625.

Also, given that the machine depreciates by 20% every year.

Therefore, its value after every year is 80% of the original cost i.e., 4/5th of the original cost.

So, the value at the end of 5 years = 15625 × (\frac{4}{5} × \frac{4}{5} ×...\hspace{0.1cm}5\hspace{0.1cm}times)  = 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years will be Rs 5120.

Question 32: 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on 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 us take x to be the number of days in which 150 workers finish the work.

According to the question, we have

150x = 150 + 146 + 142 + ….. (x + 8) terms

We know, 150, 146, 142, ….. (x + 8) terms is an A.P.

 with first term (a) = 150, 

Common difference (d) = –4 and number of terms (n) = (x + 8)

Now, we know sum of this series is equal to 150x    .

=> 150x = \frac{(x+8)}{2}    [2×150 + (x+8–1)×4]

=> 300x = (x+8)(300–4x–28)

=> 300x = 272x–4x2+2176–32x

=> 4x2+60x+–2176 = 0

=> x2+15x–544 = 0

=> x2+32x–17x-544 = 0

=> (x+32)(x–17) = 0

=> x = –32 or 17

Ignoring x = –32 as number of days is a positive quantity so, x = 17.

It took 8 more days to finish the work because workers were dropped out. 

Hence, the number of days in which the work was completed is 17+8 = 25.

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