NCERT Solutions Class 11 Chapter 8 Sequences And Series Exercise 8.2 (original) (raw)

Last Updated : 22 Apr, 2024

Question 1. Find the 20th and nth terms of the G.P 5/2, 5/4, 5/8, ...

**Solution:

According to the question

G.P: 5/2, 5/4, 5/8, ...

So, first term(a) = 5/2

So, the common ratio(r) = \frac{a_1}{a}=\frac{\frac54}{\frac52}= \frac12

Find: 20th and nth terms of the given G.P

So, the nth term of G.P can be expressed using formula:

an = arn - 1

Where a is 1st term and r is the common ratio.

Now we find the 20th terms of the given G.P:

a20 = (5/2)(1/2)20-1 = (5/2)(1/2)19= \frac{5}{2^{20}}

Find the nth terms of the given G.P:

an = \frac52\frac1{2^{n-1}}=\frac5{2^{n}}

Question 2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

**Solution:

According to the question

Common ration(r) = 2

and 8th term is 192

So let us considered a be the first term

So,

a8 = ar7

ar7 = 192

a(2)7 = 192

a = 3/2

Find: 12th term of a G.P.

As we know that the nth term of G.P can be expressed using formula:

an = arn - 1

Where a is 1st term and r is the common ratio

So, we find 12th term of a G.P.

a12 = ar12 - 1

a12 = ar11

a12 = (3/2)(2)11

a12 = 3072

Question 3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

**Solution:

According to the question

The 5th, 8th and 11th terms of a G.P. are p, q and s

Prove: q2 = ps

Proof:

Let a G.P. with first term a and common ratio r,

So, a5 = ar4 = p ....(1)

a8 = ar7= q ....(2)

a11 = ar10 = s ....(3)

Now divide eq(2) by (1), we get

ar7/ar4 = q/p

r3 = q/p ....(4)

Now divide eq(3) by (2), we get

ar10/ar7= s/q

r3 = s/q ....(5)

From eq(4) and (5), we get

q/p = s/q

q2 = ps

Hence Proved

Question 4. The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.

**Solution:

According to the question

First term(a) = – 3

and the 4th term of a G.P. is square of its second term

Find: 7th term

Let us considered r be the common ratio

As we know that the nth term of G.P can be expressed using formula:

an = arn - 1

Where a is 1st term and r is the common ratio

So, a4 = ar3

It is given that the 4th term of a G.P. is square of its second term

a4 = (a2)2

ar3 = (a2)2

ar3 = (ar)2

ar3 = a2r2

r = a

Now put the value of a, we get

r = -3

Now, we find the 7th term

a7 = ar7 - 1

a7 = ar6

= −3 x (−3)6 =−2187

Question 5. Which term of the following sequences:

(a) 2, 2√2, 4, ... is 128 ?

**Solution:

According to the question

G.P.: 2, 2√2, 4, ...

So, first term(a) = 2

So, the common ratio(r) = √2

As we know that, the nth term of G.P can be expressed using formula:

an = arn - 1

Where a is 1st term and r is the common ratio.

128 = 2(√2)n - 1

(2)7 = 2(√2)n - 1

(2)7 = 2((2)1/2)n - 1

(2)7 = 2(2)(n - 1)/2

(2)6 = (2)(n - 1)/2

6 = (n - 1)/2

12 = n - 1

12 + 1 = n

n = 13

Hence, the 13th term of the G.P. is 128

(b) √3, 3, 3√3, .... is 729 ?

**Solution:

According to the question

G.P.: √3, 3, 3√3, ....

So, first term(a) = √3

So, the common ratio(r) = √3

As we know that, the nth term of G.P can be expressed using formula:

an = arn - 1

Where a is 1st term and r is the common ratio.

729 = √3(√3)n-1

(3)6 = √3(√3)n-1

(3)6 = (3)1/2((3)1/2)n-1

(3)6 = (3)1/2(3)n-1/2

(3)6 = (3)1/2+(n-1)/2

6 = 1/2 + (n - 1)/2

6 - 1/2 = (n - 1)/2

(12 - 1)/2 = (n - 1)/2

11 = n - 1

n = 12

Hence, the 12th term of the G.P. is 729

(c) \frac13,\frac19,\frac1{27},... is \frac1{19683} ?

**Solution:

According to the question

G.P.: \frac13,\frac19,\frac1{27},...

So, first term(a) = 1/3

So, the common ratio(r) = 1/3

As we know that, the nth term of G.P can be expressed using formula:

an = arn - 1

Where a is 1st term and r is the common ratio.

\frac1{19683}=(\frac13)(\frac13)^{n-1}

(1/3)9 = (1/3)n

n = 9

Hence, the 9th term of the G.P. is 1/19683

Question 6. For what values of x, the numbers -2/7, x. -7/2... are in G.P.?

**Solution:

According to the question

Numbers are -2/7, x. -7/2...

The common ration is (r) = \frac{x}{\frac{-2}{7}} = -7x/2

Again the common ration(r) = = \frac{\frac{-7}{2}}{x} = -7/2x

So,

-7x/2 = -7/2x

7x/2 = 2/7x

14x2 = 14

x2 = 14/14

x = ±1

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

Question 7. 0.15, 0.015, 0.0015, ... 20 terms.

**Solution:

According to the question

G.P.: 0.15, 0.015, 0.0015, ...

So, first term(a) = 0.15

So, the common ratio(r) = 0.1

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(1 - r^n)}{1-r}

So, S20 = (0.15)[(1 - (0.1)20)/(1 - 0.1)]

= (0.15/0.9)(1 - (0.1)20)

= 1/6(1 - (0.1)20)

Question 8. √7​, √21​, 3√7​,... n terms.

**Solution:

According to the question

G.P.: √7​, √21​, 3√7​,...

So, first term(a) = √7​

So, the common ratio(r) = √3​

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(1 - r^n)}{1-r}

So, Sn = (√7​)[(1 - (√3)n)/(1 - √3)]

= (√7​)[(1 - (√3)n)/(1 - √3)] x [(1 + √3)/(1 + √3)]

= -(√7​)(1 + √3)/2(1 - (3)n/2)

= (√7​)(1 + √3)/2((3)n/2 - 1)

Question 9. 1, -a, a2, -a3,... n terms (if a ≠ – 1).

**Solution:

According to the question

G.P.: 1, -a, a2, -a3,...

So, first term(a) = 1

So, the common ratio(r) = -a​

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(1 - r^n)}{1-r}

So, S_n=\frac{(1)(1-(-a)^n)}{1 -(-a)}=\frac{1-(-a)^n}{1+a}

Question 10. x3, x5, x7, ... n terms (if x ≠ ± 1).

**Solution:

According to the question

G.P.: x3, x5, x7, ...

So, first term(a) = x3

So, the common ratio(r) = x2​

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(1 - r^n)}{1-r}

So,

S_n=\frac{(x^3)(1-(x^2)^n)}{1-x^2}

S_n=\frac{(x^3)(1-x^{2n})}{1-x^2}

Question 11. Evaluate \sum_{k=1}^{11} (2+3^k)

**Solution:

\sum_{k=1}^{11} (2+3^k) = (2 + 31) + (2 + 32) + (2 + 33) + .... + (2 + 311)

= (2 + 2 + ...11 terms) + (3 + 32 + 33+...11 terms)

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(1 - r^n)}{1-r}

So, Sn = 2 x 11 + 3(311 - 1)/3 - 1

Sn = 22 + 3/2((311 - 1))

Question 12. 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.

**Solution:

Let us considered the three terms a/r, a, ar of the G.P.

According to the question

\frac{a}r\times a\times ar = 1

a3 = 1

a = 1

Also.

\frac{a}{r}+a+ar=\frac{39}{10}

Now put the value of a, we get

\frac1r+1+r=\frac{39}{10}

\frac{r^2+r+1}r=\frac{39}{10}

10r2 + 10r + 10 = 39r

10r2 - 29r + 10 = 0

On solving the equation, we get

r = 2/5, 5/2

So, the G.P. is 5/2, 1, 2/5.

Question 13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

**Solution:

According to the question

G.P.: 3, 32, 33, …

So, first term(a) = 3

So, the common ratio(r) = 3

Sn = ​120

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(1 - r^n)}{1-r}

Sn = (3)(1 - (3)n)(1-3)

120 = (3)(1 - (3)n)

-240 = (3)(1 - (3)n)

-80 = 1 - (3)n

-80 - 1 = - (3)n

-81 = - (3)n

(3)4 = (3)n

n = 4

Hence, 4 terms of G.P. are needed to give the sum 120

Question 14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio, and the sum to n terms of the G.P.

**Solution:

Let us considered the first three terms of the G.P. are a, ar, ar2 and

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

Find: the first term, the common ratio, and the sum to n terms of the G.P.

According to the question

a + ar + ar2 = 16

a(1 + r + r2) = 16 .......(1)

ar3 + ar4 + ar5 = 128

ar3(1 + r + r2) = 128 .......(2)

Now divide eq(2) by (1), we get

ar3(1 + r + r2)/ a(1 + r + r2) = 128/16

r3 = 8

r = 2

Now put the value of r in eq(1), we get

a(1 + (2) + (2)2) = 16

a(1 + 2 + 4) = 16

a(7) = 16

a = 16/7

Now we find the sum to n terms of the G.P.

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(r^n-1)}{r-1}

S_n=\frac{16(2^n-1)}{7(2-1)}=\frac{16(2^n-1)}7

Question 15. Given a G.P. with a = 729 and 7th term 64, determine S7.

**Solution:

According to the question

The first term(a) = 729

and 7th term 64

Find: S7

Let the common ratio be r.

a7 = ar6 = 64

r6 = 64/729

r = ±2/3

Now we find the sum to 7th terms of the G.P.

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(r^n-1)}{r-1}

On taking r = +2/3

S_7=\frac{729((\frac23)^7-1)}{\frac23-1} = 2059

On taking r = -2/3

S_7=\frac{729((-\frac23)^7-1)}{(-\frac23-1)} = 463

Question 16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

**Solution:

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

According to the question

a + ar = -4

a5 = 4(a3)

So, ar4 = 4(ar2)

r2 = 4

r = ±2

When r = +2 then

a + ar = a + 2a = 3a = −4

a = -4/3

Then the G.P. is \frac{-4}3,\frac{-8}3,\frac{16}3,...

When r = -2 then

a + ar = a − 2a = −a = −4

a = 4

Then the G.P. is 4, -8, 16, -32,...

Question 17. If the 4th, 10th and 16th terms of a G.P. are x, y, and z, respectively. Prove that x, y, z are in G.P.

**Solution:

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

According to the question

a4 = ar3 = x ......(1)

a10 = ar9 = y ......(2)

a16 = ar15 = z ......(3)

Now divide eq(2) by (1), we get

ar9/ar3 = y /x

r6 = y /x

Divide eq(3) by (2), we get

ar15/ar9 = z/y

r6 = z/y

y /x = z/y

So x, y, z are in G.P.

Question 18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…

**Solution:

According to the question

Given Sequence: 8, 88, 888, 8888…

This sequence is neither A.P. nor G.P. but we can change it into G.P.

So, we can write as:

Sn = 8 + 88 + 888 + 8888 + … + n times

= 8(1 + 11 + 111 + 1111 + … + n times)

= 8/9(9 + 99 + 999 + 9999 + … + n times)

= 8/9((10 - 1) + (102 - 1) + (103 - 1) + (104 - 1) + … + n times)

= 8/9((10 + 102 + 103 + 104 + … + n times) - (1 + 1 + 1 + 1 + ....+ n terms))

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by

S_n=\frac{a(r^n-1)}{r-1}

= \frac{8}{9}[\frac{10(10^n - 1)}{10 - 1}-n]

= \frac{8}{9}[\frac{(10^{n+1} - 10)}{9}-n]

S_n=\frac{80(10^n-1)}9-\frac{8n}9

Question 19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2, 1/2.

**Solution:

According to the question

Sequence 1: 2, 4, 8, 16, 32

Sequence 2: 128, 32, 8, 2, 1/2

Product of corresponding terms are

= 2 x 128, 4 x 32, 8 x 8, 16 x 2, 32 x 1/2

= 256, 128, 64, 32, 16

So, the first term(a) = 256

Common ration = 1/2

S5 = 256[1 - (1/2)5]/1/2

= 496

Question 20. Show that the products of the corresponding terms of the sequences a, ar, ar2 , …arn – 1 and A, AR, AR2, … ARn – 1 form a G.P, and find the common ratio.

**Solution:

According to the question

Sequence 1: a, ar, ar2 , …arn – 1

Sequence 2: A, AR, AR2, … ARn – 1

Prove: aA, arAR, ar2AR2, .... arn – 1ARn – 1 from G.P

Now we find the common ration of the G.P

r = arAR. aA = rR

Again,

r = ar2AR2/arAR = rR

Hence, the sequence form G.P. and the common ratio is rR

Question 21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

**Solution:

Let us considered a be the first term, r be the common ratio and

four numbers in G.P. are a, ar, ar2,ar3

According to the question

So, a3 = a1 + 9

ar2 = ar + 9

ar2 - ar = 9 ....(1)

a2 = a4 + 18

ar = ar3 + 18

ar - ar3 = 18 ....(2)

Now divide eq(2) by (1), we get

ar - ar3/ar2 - ar = 18/9

ar (1 - r2)/-a(1 - r2) = 2

ar/-a = 2

r = -2

Now put the value of r in eq(1), we get the value of a

a(-2)2 - a(-2) = 9

a = 3

Hence, the four numbers in G.P. are 3, 3(-2), 3(-2)2,3(-2)3

= 3, -6, 12, -24.

Question 22. If the p th, qth, and rth terms of a G.P. are a, b and c, respectively. Prove that aq–rbr–pcp–q = 1.

**Solution:

Let us considered the 1st term of a G.P. be k and common ratio x.

According to the question

kxp-1 = a ....(1)

kxq-1 = b ....(2)

kxr-1 = c ....(3)

Prove: aq–rbr–pcp–q = 1

Proof:

Lets take L.H.S

i.e., aq–rbr–pcp–q

Now put the value of a, b, c from the above equations,

aq – rbr – pcp – q = (kxp - 1)q – r(kxq - 1)r – p(kxr - 1)p – q

= k0 x x0

= 1

L.H.S = R.H.S

Question 23. If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.

**Solution:

According to the question

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

Let the G.P. are a, ar, ar2,ar3. Here, r be the common ratio.

Then b = arn-1

Now, Product of n terms(P) = a x ar x ar2 x ... x arn-1

P = anr{1 + 2 + ...n - 1}

P = anr{(n(n - 1)/2}

So, P2 = a2nr(n(n - 1)

= [a2r(n - 1)]n

= [a x ar(n - 1)]n

= [ab]n

Hence Proved.

Question 24. 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 the first term of G.P. be a and the common ratio be r.

So, the sum of 1st n terms is S_{1n}=\frac{a(r^n-1)}{r-1}

According to the question

Sum of (n+1)th to (2n)th term is S_{2n}=\frac{ar^n(r^n-1)}{r-1}

The required ratio is:

\frac{S_{1n}}{S_{2n}}=\frac{\frac{a(r^n-1)}{r-1}}{\frac{ar^n(r^n-1)}{r-1}} =\frac{a(r^n-1)}{r-1} \times\frac{r-1}{ar^n(r^n-1)} = \frac{1}{r^n}

Question 25. If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2

**Solution:

According to the question

a, b, c, d are in G.P., so let the common ratio of G.P.be r.

Then b = ar, c = ar2, d = ar3

Simplifying LHS by putting value of b, c, d

(a2 + b2 + c2)(b2 + c2 + d2) = (a2 + a2r2 + a2r4)(a2r2 + a2r4 + a2r6)

= a4r2(1 + r2 + r4)2

Now, simplifying RHS

(ab + bc + cd)2 = (a2r + a2r3 + a2r5)2 = a4r2(1 + r2 + r4)2

LHS = RHS

Question 26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

**Solution:

Let us considered x1 and x2 be the two numbers in between 3 and 81. So, the G.P. is 3, x1, x2, 81 and

r be the common ratio. First term of the G.P.(a) = 3

So, a4 = 81

(3)r3 = 81

r3 = 27

r = 3

So, x1 = ar = 3 x 3 = 9

x2 = ar2 = (3)(3)2 = 27

Hence, the G.P. is 3, 9, 27, 81

Question 27. Find the value of n so that \frac{a^{n+1}+b^{n+1}}{a^n+b^n} may be the geometric mean between a and b.

**Solution:

G.M. between a and b is √ab

\sqrt{ab}=\frac{a^{n+1}+b^{n+1}}{a^n+b^n}

On squaring both side we get

ab=\frac{(a^{n+1}+b^{n+1})^2}{(a^n+b^n)^2}

ab(an + bn)2 = (an + 1 + bn + 1)2

ab(a2n + b2n + 2anbn) = a2n + 2 + b2n + 2 + 2a2n + 2 b2n + 2

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

ba2n+1 + ab2n+1 = a2n + 2 + b2n + 2

ab2n+1 - b2n + 2 = a2n + 2 - ba2n+1

ab2n+1 - b2n + 2 = a2n + 2 - ba2n+1

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

b2n+1 = a2n + 1

a2n+1 / b2n + 1 = 1 = (a/b)0

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

2n + 1 = 0

n = -1/2

Question 28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+2\sqrt2):(3-2\sqrt2)

**Solution:

Let the two numbers be a and b.

And G.M. = √ab

According to the question

a + b = 6√ab .....(1)

Squaring on both side, we get

(a + b)2 = 36(ab)

As we know that

(a - b)2 = (a + b)2 - 4(ab)

So,

(a - b)2 = 36(ab) - 4(ab)

a - b = 4√2√ab ......(2)

Now we add eq(a) and (2), we get

a + b + a - b = 6√ab + 4√2√ab

2a = √ab(6 + 4√2)

a = √ab(3 + 2√2)

Now put the value of a in eq(1), we get the value of b

√ab(3 + 2√2) + b = 6√ab

b = √ab(3 - 2√2)

Now we find the ratio:

a/b = √ab(3 + 2√2)/√ab(3 - 2√2)

= (3 + 2√2)/(3 - 2√2)

Question 29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A\pm\sqrt{(A+G)(A-G)}

**Solution:

Let the two numbers be a & b.

A.M = A= (a + b)/2

G.M = G = 2/√ab

So, a + b = 2A ....(1)

G2 = ab ....(2)

As we know that

(a - b)2 = (a + b)2 - 4(ab)

So,

(a - b)2 = (2A)2 - 4(G2)

(a - b)2 = 4(A)2 - 4(G2)

(a - b)2 = 4[(A)2 - (G2)]

(a - b)2 = 4[(A - G)(A + G)]

(a - b) = √4[(A - G)(A + G)]

(a - b) = 2√(A - G)(A + G) ......(3)

Now add eq(1) and (3), we get

2a = 2A + 2√(A - G)(A + G)

a = A + √(A - G)(A + G)

Now put the value of a in eq(1), we get

A + √(A - G)(A + G) + b = 2A

b = A - √(A - G)(A + G)

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

Question 30. 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:

As the count of bacteria doubles after each hour, so at the end of n hours it becomes 2n times the original count.

So, first term(a) = 30, and common ratio(r) = 2

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

At the end of 2nd hour there are 120 bacteria.

a4 = ar4 = (30)(2)4 = 480

At the end of 4nd hour there are 480 bacteria.

an+1 = arn = (30)(2)n

At the end of nnd hour there are (30)(2)n bacteria.

Question 31. What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays an annual interest rate of 10% compounded annually?

**Solution:

In the end of 1st year, the amount = 500(1 + 1/10) = 500(1.1)

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

In the end of 3rd year, the amount = 500(1.1)(1.1)(1.1)

.....So on

At the end of 10 year, the amount = 500(1.1)(1.1)(1.1).......10 times

= 500(1.1)10

Question 32. 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 roots of the quadratic equation be a and b.

According to the question

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

a + b = 16

G.M. = √ab = 5

ab = 25

Then the quadratic equation using the roots can be written as

x2 - (a + b)x + ab = 0

Now put all these values in the quadratic equation, we get

x2 - 16x + 25 = 0