**Q1. ****Find 20 ^{th} and n^{th }term for the G.P **

**Soln:**

Given G.P =

Here, first term = a =

Common ratio = r =

**Q2. Find 12 ^{th} term for the G.P that has 8^{th }term 192 and common ratio of 2.**

**Soln:**

Given,

Common ratio = r = 2

Assume = first term = a

**Q3. If p is the 5 ^{th}, q is the 8^{th} and s is the 11^{th} term of a G.P. Prove that q^{2} = ps**

**Soln:**

Let the first term be ‘a’ and the common ratio be r for the G.P.

Given condition

Eqn (2) divided by Eqn (1), we have

Eqn (3) divided by Eqn (2), we have

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

Hence proved

**Q4. If first term of a G.P is -3 and the 4 ^{th} term is square of the 2^{nd} term. Find the 7^{th} term.**

**Soln:**

Let the first term be ‘a’ and common ratio be ‘r’.

Therefore, a = -3

We know that, a_{n} = ar^{n – 1}

a_{4} = ar^{3} = (-3)r^{3}

a_{2} = ar^{1} = (-3)r

Given condition

(-3)r^{3} = [(-3) r]^{2}

r^{6} = (-3) (-3)^{6} = – (3)^{7} = -2187

Hence, the 7^{th} term is -2187.

**Q5. Which term of**

**(a) 2,22–√,4,....is128?**

**(b) 3–√,3,33–√....is729?**

**(c) 13,19,127,....is119683?**

**Soln:**

(a) Given sequence =

Here,

First term = a = 2

Common ratio = r =

Assume n^{th }term = 128

Hence, 128 is the 13^{th }term

(b) Given sequence =

Here,

First term = a =

Common ratio = r =

Assume n^{th }term = 729

Therefore,

Hence, 729 is the 12^{th }term of the sequence.

(c) Given sequence =

Here,

First term = a =

Common term = r =

Assume the nth term be

n = 9

Hence, ^{th }term of the sequence.

**Q6. If 27,x,−72 are in G.P then find the value of x?**

**Soln:**

Given sequence =

Common ratio =

Also,

Hence, the sequence is in G.P if

**Q7. The sum of the first 20 terms of the G.P 0.15, 0.015, 0.0015 . . . .**

**Soln:**

Given G.P = 0.15, 0.015, 0.0015 . . . .

Here,

First term = a = 0.15

Common ration = r =

**Q8. 7–√,21−−√,37–√,.... is a G.P series. Find sum till the n ^{th }term.**

**Soln:**

Given series =

Here,

First term = a =

Common ratio = r =

**Q9. What is the sum of the n ^{th }term of the G.P series**

**1, -a, a ^{2}, -a^{3} . . . . (if a ≠ -1)?**

**Soln:**

Given series = 1, -a, a^{2}, -a^{3} . . .

Here,

The first term of the G.P = a_{1} = 1

Common ratio of the G.P = r = -a

**Q10. What is the sum of the n ^{th }term of the G.P series**

**x ^{3}, x^{5}, x^{7 }. . . . (if x ≠ ± 1)**

**Soln:**

Given series = x^{3}, x^{5}, x^{7 }. . . .

Here,

First term = a = x^{3}

Common ratio = r = x^{2}

**Q11. Evaluate ∑11k=1(2+3k)**

**Soln:**

The above sequence is in a G.P 3, 3^{2}, 3^{3}, . . .

Substituting eqn (2) in eqn (1), we get

**Q12. Sum of 1 ^{st} 3 terms in a G.P is 3910 and the product of the terms is 1. What is the the terms and common ratio of the sequence?**

**Soln:**

Assume ^{st} 3 terms in the G.P

From eqn (2), we get a^{3 }= 1

a = 1

Substituting (a = 1) in eqn (1), we get

Hence,

The terms are

**Q13. For the G.P 3, (3) ^{2}, (3)^{3 }…. sum of how many terms is 120?**

**Soln:**

Given G.P = 3, (3)^{2}, (3)^{3 }….

Let n number of terms are required for obtaing the sum of 120

Here,

First term = a = 3

Common ratio = r = 3

n = 4

Hence, 4 terms are required to get the sum of 120

**Q14. In a G.P sum of the first 3 terms of are 16 where as sum for the next 3 terms are 128.**

**Find common ratio, first term and sum of n terms for the G.P**

**Soln:**

Assume G.P = a, ar, a(r)^{2}, a(r)^{3}, . . .

Given condition =

a + a(r)^{2 }+ ar + = 16 and a(r)^{5} + a(r)^{4 }+ a(r)^{3} = 128

a (1 + (r)^{2 }+ r) = 16 …… (1)

a(r)^{3}(1+ (r)^{2} + r) = 128 …….. (2)

eqn (2) divided by eqn (1), we get

r^{3} = 8

r = 2

substituting r = 2 in eqn(1), we get

a(1 + 2 + 4) = 16

a(7) = 16

**Q15. Given first term ‘a’ = 729 and the 7 ^{th }term = 64, then determine S_{7}.**

**Soln:**

Given a = 729 , a_{7} = 64

Assume common ratio = r

We know that,

a_{n} = a r^{n – 1}

a_{7} = a r^{7 – 1} = (729) (r)^{6}

64 = 729 (r)^{6}

Also, we know that

= 2187 – 128

= 2059

**Q16. The sum of first 2 terms of a G.P is -4 and 5 ^{th} term is 4 times that of third term. Find the G.P**

**Soln:**

Assume first term = a

Common ratio = r

From (1), we get

-4 = a(3)

Also,

a = 4

Hence, required G.P is

**Q17. If p, q, and s are fourth, tenth and sixteenth terms of G.P prove that p, q, s are in G.P**

**Soln:**

Let

First term = a

Common ratio = r

According to given condition,

a_{4 }= ar^{3} = p ………. (1)

a_{10 }= ar^{9} = q ………. (2)

a_{16 }= ar^{15} = x ………. (3)

eqn (2) divided by eqn (1)

eqn (3) divided by eqn (2)

**Q18. 8, 88, 888, 8888, …… are in sequence find sum of the n ^{th} terms**

**Soln:**

Given sequence 8, 88, 888, 8888, ……

This sequence isn’t a G.P

It may be changed to a G.P if the terms are written as S_{n }= 8 + 88 + 888 + 8888 + ….. to n terms

**Q19. If 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½ are in sequences find the sum of the products**

**Soln:**

Required sum =

Here,

Now first term, a = 4

And common ratio,

We know that,

Required sum =

**Q20. Prove that product of corresponding terms from a, ar, ar ^{2},_{… }ar^{n-1} and A, AR, AR^{2},_{… }AR^{n-1} are in sequences find common ratio**

**Soln:**

To show = aA, ar(AR), ar^{2 }(AR^{2}), … ar^{n – 1 }(AR^{n – 1}) is a G.P series

Hence, the sequences are in G.P and rR is common ratio.

**Q21. If in a G.P of 4 numbers 3 ^{rd }term is greater to 1^{st} term by 9, and the 2^{nd} term is greater to 4^{th} by 18.**

**Soln:**

Let

First term = a

Common ratio = r

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

From given conditions,

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

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

From eqn (1) and eqn (2), we get

a(r^{2} – 1) = 9 ……… (3)

ar(1 – r^{2}) = 18 ……..(4)

Dividing eqn (4) by eqn (3), we get

=> – r = 2

=> r = -2

Substituting r = -2 in eqn (1), we get

4a = a + 9

=> 3a = 9

=> a = 3

Hence, first four terms of G.P are 3, 3(-2), 3(-2)^{2}, and 3(-2)^{3}

i.e. 3, -6, 12, -24

**Q22. If in a G.P a, b and c are the p ^{th}, q^{th} and r^{th} terms. Prove that**

**a ^{q – r}.b^{r – p}.c^{p – q }= 1**

**Soln:**

Let

First term = a

Common ratio = r

From the given condition

AR^{p – 1} = a

AR^{q – 1} = b

AR^{r – 1} = c

a^{q – r}.b^{r – p}.c^{p – q}

= A^{0} x R^{0}

= 1

Hence proved

**Q23. If in a G.P ‘a’ and ‘b’ are the first and n ^{th} term and product of n^{th} terms is P.**

**Show P ^{2 }= (ab)^{n}**

**Soln:**

Given:

First term = a

Last term = b

Therefore,

G.P = 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 x a x …. a)(r x r^{2} x … r^{n – 1})

= an r 1 + 2 + … (n – 1) …… (2)

Here, 1, 2, …..(n – 1) is an A.P

1 + 2 + …. + (n – 1)

= (ab)^{n}

Hence proved

**Q24. Prove that the ratio of sum of first n terms to sum of terms from (n + 1) ^{th} to (2n)^{th }terms is 1/ r^{n} .**

**Soln:**

Let

First term = a

Common ratio = r

Sum of the first n terms =

Since there are ‘n’ number of terms from (n + 1)^{th }to (2n)^{th }term,

Sum of the terms

Required ratio =

Thus, the ratio is 1/r^{n}

**Q25. Prove that:**

**(a ^{2} + b^{2 }+ c^{2}) (b^{2} + c^{2 }+ d^{2}) = (ab + bc – cd)^{2}**

**If a, b, c and d are in a G.P series**

**Soln:**

a, b, c and d are in a G.P series

Therefore

bc = ad …………….. (1)

b^{2} = ac ……………… (2)

c^{2} = bd ……………… (3)

It is to be proven 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 } [From (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 } [From (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} x b^{2 }+ b^{2}c^{2 }+ b^{2}d^{2} + c^{2}b^{2} + c^{2} x c^{2} + c^{2}d^{2
}[From eqn (1) and (2)]

= 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

**Q26. Between 3 and 81 insert two numbers such that the sequence is a G.P series.**

** Soln:**

Let the numbers be g_{1 }and g_{2}

So from the given condition 3, g_{1,} g_{2}, 81 is in G.P

Assume first term = a

Common ratio = r

Therefore 81 = (3)(r)^{3}

=> r^{3} = 27

r = 3 (taking real roots only)

For r = 3,

Q26.

G_{1} = ar = (3)(3) = 9

G_{2} = ar^{2} = (3)(3)^{2} = 27

Hence, the numbers are 9 and 27

**Q27. If a2+1+bn+1an+bn is the geometric mean for a and b. What is the value of n?**

**Soln:**

Mean of a and b is

From the given condition

Squaring both the sides, we get

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

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

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

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

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

=>

=> 2n + 1 = 0

=>

**Q28. Prove that the ratio of two numbers is (3+22–√)(3–22–√) if the sum of the two numbers is 6 times their geometric mean.**

**Soln:**

Let the two numbers be a and b.

G.M =

From given conditions,

=> (a + b)^{2} = 36(ab)

Also,

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

=>

=

Eqn(1) + Eqn (2)

=>

Putting the value of a in eqn (1),we get

=>

Hence, required ratio =

**Q29. For two positive numbers if G and A are the G.M and A.M show that the numbers are A±(A+G)(A–G)−−−−−−−−−−−−√**

**Soln:**

Given G and A are A.M and G.M

Assume the numbers be a and b.

From Eqn(1) and Eqn (2), we get

a + b = 2A ……. (3)

ab = G^{2} ……. (4)

Putting the value of a and b from (3) and (4)

(a – b)^{2} = (a + b)^{2} – 4ab, we get

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

From eqn (3) and (5), we get

=>

Putting the value of a in eqn(3), we get

Hence, the numbers are

**Q30. Bacteria over certain culture multiply twice in one hour. If in initial stage there were only 30 bacteria, what will be the amount of bacteria in the 2 ^{nd}, 4^{th} and n^{th} hour?**

**Soln:**

Given that amount of bacteria multiplies two times in one hour

Therefore,

Bacteria amount will form G.P

Here, first term = (a) = 30

Common ratio = (r) = 2

Therefore ar^{2} = a_{3 }= 30 x (2)^{2 }= 120

Therefore amount of bacteria after 2^{nd} hour is 120

ar^{4} = a_{5 }= 30 x (2)^{4 }= 480

Therefore amount of bacteria after 4^{th} hour is 480

_{ }ar^{n }x a_{n + 1} = 30 x (2)^{n}

Hence, amount of bacteria after n^{th} hour is 30 x (2)^{n}