# Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.3 | Set 2

### Question 17. If the 4^{th}, 10^{th} and 16^{th} 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

a

^{4}= ar^{3 }= x ……(1)a

_{10}= ar^{9}= y ……(2)a

_{16}= ar^{15}= z ……(3)Now divide eq(2) by (1), we get

ar

^{9}/ar^{3 }= y /xr

^{6 }= y /xDivide eq(3) by (2), we get

ar

^{15}/ar^{9 }= z/yr

^{6 }= z/yy /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:

S

_{n }= 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) + (10

^{2 }– 1) + (10^{3 }– 1) + (10^{4 }– 1) + … + n times)= 8/9((10 + 10

^{2 }+ 10^{3 }+ 10^{4 }+ … + n times) – (1 + 1 + 1 + 1 + ….+ n terms))As we know that, the sum of n terms of G.P. with 1

^{st}term a & common ratio r is given by=

=

### 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

S

_{5 }= 256[1 – (1/2)^{5}]/1/2= 496

### Question 20. Show that the products of the corresponding terms of the sequences a, ar, ar^{2 }, …ar^{n – 1 }and A, AR, AR^{2}, … AR^{n – 1 }form a G.P, and find the common ratio.

**Solution:**

According to the question

Sequence 1: a, ar, ar

^{2 }, …ar^{n – 1}Sequence 2: A, AR, AR

^{2}, … AR^{n – 1}Prove: aA, arAR, ar

^{2}AR^{2}, ….^{ }ar^{n – 1}AR^{n – 1}from G.PNow we find the common ration of the G.P

r = arAR. aA = rR

Again,

r = ar

^{2}AR^{2}/arAR = rRHence, 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, ar

^{2},ar^{3}According to the question

So, a

_{3}= a_{1}+ 9ar

^{2}= ar + 9ar

^{2 }– ar = 9 ….(1)a

_{2}= a_{4}+ 18ar = ar

^{3}+ 18ar – ar

^{3}= 18 ….(2)Now divide eq(2) by (1), we get

ar – ar

^{3}/ar^{2 }– ar = 18/9ar (1 – r

^{2})/-a(1 – r^{2}) = 2ar/-a = 2

r = -2

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

a(-2)

^{2 }– a(-2) = 9a = 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}, q^{th},^{ }and r^{th} terms of a G.P. are a, b and c, respectively. Prove that a^{q–r}b^{r–p}c^{p–q }= 1.

**Solution:**

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

According to the question

kx

^{p-1 }= a ….(1)kx

^{q-1 }= b ….(2)kx

^{r-1 }= c ….(3)Prove: a

^{q–r}b^{r–p}c^{p–q }= 1Proof:

Lets take L.H.S

i.e., a

^{q–r}b^{r–p}c^{p–q }Now put the value of a, b, c from the above equations,

a

^{q – r}b^{r – p}c^{p – q }= (kx^{p – 1})^{q – r}(kx^{q – 1})^{r – p}(kx^{r – 1})^{p – q }= k

^{0}x x^{0}= 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 P^{2} = (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, ar

^{2},ar^{3}. Here, r be the common ratio.Then b = ar

^{n-1}Now, Product of n terms(P) = a x ar x ar

^{2 }x … x ar^{n-1}P = a

^{n}r^{{1 + 2 + …n – 1} }P = a

^{n}r^{{(n(n – 1)/2} }So, P

^{2}= a^{2n}r^{(n(n – 1) }= [a

^{2}r^{(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/r^{n}.

**Solution:**

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

So, the sum of 1st n terms is

According to the question

Sum of (n+1)

^{th}to (2n)^{th}term isThe required ratio is:

### Question 25. If a, b, c and d are in G.P. show that (a^{2 }+ b^{2 }+ c^{2})(b^{2 }+ c^{2 }+ d^{2}) = (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 = ar

^{2}, d = ar^{3 }Simplifying LHS by putting value of b, c, d

(a

^{2 }+ b^{2 }+ c^{2})(b^{2 }+ c^{2 }+ d^{2}) = (a^{2 }+ a^{2}r^{2 }+ a^{2}r^{4})(a^{2}r^{2 }+ a^{2}r^{4 }+ a^{2}r^{6})= a

^{4}r^{2}(1 + r^{2 }+ r^{4})^{2}Now, simplifying RHS

(ab + bc + cd)

^{2 }= (a^{2}r + a^{2}r^{3 }+ a^{2}r^{5})^{2 }= a^{4}r^{2}(1 + r^{2 }+ r^{4})^{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, a

^{4 }= 81(3)r

^{3}= 81r

^{3}= 27r = 3

So, x1 = ar = 3 x 3 = 9

x2 = ar

^{2}= (3)(3)^{2}= 27Hence, the G.P. is 3, 9, 27, 81

### Question 27. Find the value of n so that may be the geometric mean between a and b.

**Solution:**

G.M. between a and b is √ab

On squaring both side we get

ab(a

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

^{2n}+ b^{2n}+ 2a^{n}b^{n}) = a^{2n + 2}+ b^{2n + 2}+ 2a^{2n + 2}b^{2n + 2}a

^{2n+1}b + ab^{2n+1}+ 2a^{n+1}b^{n+1}= a^{2n + 2}+ b^{2n + 2}+ 2a^{2n + 2}b^{2n + 2}ba

^{2n+1}+ ab^{2n+1 }= a^{2n + 2}+ b^{2n + 2}ab

^{2n+1 }– b^{2n + 2 }= a^{2n + 2 }– ba^{2n+1}ab

^{2n+1 }– b^{2n + 2 }= a^{2n + 2 }– ba^{2n+1}b

^{2n+1 }(a – b) = a^{2n + 1}(a – b)b

^{2n+1 }= a^{2n + 1}a

^{2n+1 }/ b^{2n + 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

**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

**Solution:**

Let the two numbers be a & b.

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

G.M = G = 2/√ab

So, a + b = 2A ….(1)

G

^{2}= ab ….(2)As we know that

(a – b)

^{2}= (a + b)^{2 }– 4(ab)So,

(a – b)

^{2}= (2A)^{2 }– 4(G^{2})(a – b)

^{2}= 4(A)^{2 }– 4(G^{2})(a – b)

^{2}= 4[(A)^{2 }– (G^{2})](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 = 2Ab = 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 2^{nd} hour, 4^{th} hour, and n^{th} hour?

**Solution:**

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

^{n}times the original count.So, first term(a) = 30, and common ratio(r) = 2

So, a

_{3}= ar^{2}= (30)(2)^{2}= 120At the end of 2

^{nd}hour there are 120 bacteria.a

_{4}= ar^{4}= (30)(2)^{4}= 480At the end of 4

^{nd}hour there are 480 bacteria.a

_{n+1}= ar^{n}= (30)(2)^{n}At the end of n

^{nd}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

x

^{2 }– (a + b)x + ab = 0Now put all these values in the quadratic equation, we get

x

^{2 }– 16x + 25 = 0

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