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

Solution:

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

a_{1} = a, a_{2} = ar, a_{3} = ar^{2}, a_{4} = ar^{3}

By the given condition,

a_{3} = a_{1} + 9 ⇒ ar^{2} = a + 9 ………………. (1)

a_{2} = a_{4} + 18 ⇒ ar = ar^{3} + 18 ………… (2)

From (1) and (2), we obtain

a(r^{2} – 1) = 9 …………………….. (3)

ar (1– r^{2}) = 18 …………………. (4)

Dividing (4) by (3), we obtain

[ar(1-r^{2})]/[a(r^{2}-1)] = ^{18}/_{9}

-r = 2

r = -2

Substituting the value of r in (1), we obtain

4a = a + 9

⇒ 3a = 9

∴ a = 3

Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)^{2}, and 3(–2)^{3} i.e., 3 ̧–6, 12, and –24.

**If p**a^{th}, q^{th}and r^{th}terms of a G.P. are a, b and c respectively. Prove that^{q-r}. b^{r-p}.C^{p – q}= 1Solution:Let A be the first term and R be the common ratio of the G.P.According to the given information,

AR

^{p-1 }= aAR

^{q-1 }= bAR

^{r-1 }= ca

^{q-r}. b^{r-p}.C^{p – q}= A^{q-r }x R^{(p-1)(q-r) }x A^{r-p }x R^{(q-1)(r-p) }x A^{p – q}x R^{(r-1)(p-q)}=A

^{q-r+r-p+p-q }x R^{(pr – pr – q –pq) + (rq – r + p – pq) + (pr – p – qr + q)}= A

^{0}xR^{0}= 1

Thus, the given result is proved.

**If the first and the nth term of G.P. are a and b, respectively, and if p is the product of n terms, prove that p**^{2}= (ab)^{n}

Solution:

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

Therefore, the G.P. is a, ar, ar

^{2}, ar^{3}… ar^{n–1}, where r is the common ratio.b = ar

^{n–1}……………………………. (1)P = Product of n terms

= (a) (ar) (ar

^{2}) … (ar^{n–1})= (a × a ×…a) (r × r

^{2}× …r^{n–1})= an r

^{1 + 2 +…(n–1)}… (2)Here, 1, 2, …(n – 1) is an A.P.

1 + 2 + ……….+ (n – 1) =

^{(n-1)}/_{2}[2 + (n-1-1) x 1] =^{(n-1)}/_{2}[2 + n – 2] =^{n(n-1)}/_{2=}P = a

^{n}r^{[n(n-1)/2]}P

^{2}= a^{2n}. r^{n(n-1)}= [a

^{2}.r^{(n-1)}]^{n}=[ a x ar

^{n-1}]^{n}= (ab)

^{n}Thus, the given result is proved.

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

Sum o first n terms = [a(1-r

^{n})]/(1-r)Since there are n terms from (n+1)

^{th}to (2n)^{th}term,Sum of terms from (n +1)

^{th}to (2n)^{th}termThus, 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}.**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:

a, b, c, d are in G.P. Therefore,

bc = ad …………………… (1)

b

^{2}= ac …………………… (2)c

^{2}= bd ………………….. (3)It has to be proved that,

(a

^{2}+ b^{2}+ c^{2}) (b^{2}+ c^{2}+ d^{2}) = (ab + bc – cd)^{2}R.H.S.

= (ab + bc + cd)

^{2}= (ab + ad + cd)

^{2}[Using (1)]= [ab + d (a + c)]

^{2}= a

^{2}b^{2}+ 2abd (a + c) + d^{2}(a + c)^{2}= a

^{2}b^{2}+2a^{2}bd + 2acbd + d^{2}(a^{2}+ 2ac + c^{2})= a

^{2}b^{2}+ 2a^{2}c^{2}+ 2b^{2}c^{2}+ d^{2}a^{2}+ 2d^{2}b^{2}+ d^{2}c^{2}[Using (1) and (2)]= a

^{2}b^{2}+ a^{2}c^{2}+ a^{2}c^{2}+ b^{2}c^{2}+ b^{2}c^{2}+ d^{2}a^{2}+ d^{2}b^{2}+ d^{2}b^{2}+ d^{2}c^{2}= a

^{2}b^{2}+ a^{2}c^{2}+ a^{2}d^{2}+ b^{2}× b^{2}+ b^{2}c^{2}+ b^{2}d^{2}+ c^{2}b^{2}+ c^{2}× c^{2}+ c^{2}d^{2}[Using (2) and (3) and rearranging terms]

= a

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

^{2}+ b^{2}+ c^{2}) (b^{2}+ c^{2}+ d^{2}) = L.H.S.∴ L.H.S. = R.H.S.

∴ (a

^{2}+ b^{2}+ c^{2}) (b^{2}+ c^{2}+ d^{2}) = (ab + bc – cd)^{2}_{ }**Insert two numbers between 3 and 81 so that the resulting sequence is G.P.**

Solution:

Let G

_{1}and G_{2}be two numbers between 3 and 81 such that the series, 3,G

_{1}, G_{2}, 81, forms a G.P.Let a be the first term and r be the common ratio of the G.P.

∴ 81 = (3) (r)

^{3}⇒ r

^{3}= 27∴ r = 3 (Taking real roots only)

For r = 3,

G

_{1}= ar = (3) (3) = 9G

_{2}= ar^{2}= (3) (3)^{2}= 27Thus, the required two numbers are 9 and 27.

**Find the value of n so that [a**^{n+1}+b^{n+1}]/[a^{n}+b^{n}] may be the geometric mean between a and b.

Solution:

- of a and b is √ab

By the given condition [a

^{n+1}+b^{n+1}]/[a^{n}+b^{n}] = √abSquaring both sides, we obtain

[a

^{n+1}+b^{n+1}]^{2}/[a^{n}+b^{n}]^{2}= aba

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

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

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

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

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

^{a}/_{b})^{2n+1}= 1 = (^{a}/_{b})^{0}2n + 1 = 0

n = –

^{1}/_{2}**The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2)( 3 – 2√2)**

Solution:

Let the two numbers be a and b.

G.M. = √(ab)

According to the given condition,

a + b = 6√(ab) ———(1)

(a+b)

^{2}= 36(ab)Also, (a –b)

^{2}= (a + b)^{2}– 4ab = 36ab – 4ab = 32 aba – b = √32. √(ab)

=4√2. √(ab) ———(2)

Adding (1) and (2) , we get

2a = (6 + 4√2) √(ab)

a = (3 + 2√2) √(ab)

Substituing the value of a in (1), we get

b = 6√(ab) – (3+2√2) √(ab)

b = (3 – 2√2) √(ab)

^{a}/_{b}=^{[(3 + 2√2) √(ab)]}/_{[ (3 – 2√2) √(ab)]}=^{3 + 2√2}/_{3 + 2√2}Thus, the required ratio is (3 + 2√2):(3 – 2√2)

**If A and G be A.M. and G.M. respectively between two positive numbers prove that the numbers are A ± √(A+G)(A-G)**

Solution:

It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.

A.M. = A =

^{a+b}/_{2}————–(1)GM = G = √(ab) ——————-(2)

From (1) and (2), we get

a + b = 2A ————-(3)

ab = G

^{2}—————(4)Substituting the value of a and b from (3) and (4) I the identity

(a – b)

^{2}= (a + b)^{2}– 4ab, we get(a – b)

^{2}= 4A^{2}– 4G^{2}= 4(A^{2}– G^{2})(a – b)

^{2}= 4(A+G)(A-G) ————-(5)From (3) and (5), we get

2a = 2A + 2√[(A+G)(A – G)]

a = A + √[(A+G)(A – G)]

substituting the value of a in (3), we obtain

b = 2A – A – √(A+G)(A-G) = A – √(A+G)(A-G)

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

- T
**he 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:

It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P.

Here, a = 30 and r = 2

a

_{3}= ar^{2}= (30) (2)^{2}= 120Therefore, the number of bacteria at the end of 2

^{nd}hour will be 120.a

_{5}= ar^{4}= (30) (2)^{4}= 480The number of bacteria at the end of 4th hour will be 480.

a

_{n +1}= ar^{n}= (30)2^{n}Thus, number of bacteria at the end of n

^{th}hour will be 30(2)^{n}.**What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?**

Solution:

The amount deposited in the bank is Rs 500.

At the end of first year, amount = Rs 500 (1+

^{1}/_{10}) = Rs. 500(1.1)At the end of 2nd year, amount = Rs 500 (1.1) (1.1)

At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on

Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)

^{10}**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 root of the quadratic equation be a and b.

According to the given condition,

A.M. =

^{a+b}/_{2}= 8a + b = 16 ——–(1)

G.M. = √(ab) = 5

ab = 25 ————-(2)

The quadratic equation is given by,

x

^{2}– x (Sum of roots) + (Product of roots) = 0x

^{2}– x (a + b) + (ab) = 0x

^{2}– 16x + 25 = 0 [Using (1) and (2)]Thus, the required quadratic equation is x

^{2}– 16x + 25 = 0

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