# 10th Maths -Chapter 2- Numbers and Sequences -Exercise 2.7 (All Sums)

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### 10th Maths -Chapter 2- Numbers and Sequences -Exercise  2.7 (All Sums)

Question 1.

Which of the following sequences are in G.P?

(i) 3, 9, 27, 81, ……..

(ii) 4, 44, 444, 4444, ………

(iii) 0.5, 0.05, 0.005, ……..

(iv) 1/3,1/6,1/12………

(v) 1, -5, 25, -125, …….

(vi) 120, 60, 30, 18, …….

(vii) 16, 4, 1, 1/4, ……

Solution:

(i) 3, 9, 27, 81

r = Common ratio

Question 2.

Write the first three terms of the G.P. whose first term and the common ratio are given below.

(i) a = 6, r = 3

(ii) a = √2 , r = √2

(iii) a = 1000, r = 2/5

Solution:

(i) a = 6, r = 3

tn = arn-1

t1 = ar1-1 = ar0 = a = 6

t2 = ar2-1 = ar1 = 6 × 3 = 18

t3 = ar3-1 = ar2 = 6 × 32 = 54

∴ The 3 terms are 6, 18, 54, ….

The 3 terms are 1000, 400, 160, ……………

Question 3.

In a G.P. 729, 243, 81,… find t7.

Solution:

G.P = 729, 243, 81 ……

t7 = ?

Question 4.

Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.

Solution:

t2/t1 = x+12/x+6,t3/t2 = x+15/x+12

Since it is a G.P.

x+12/x+6 = x+15/x+12

(x + 12)2 = (x + 6) (x + 15)

x2 + 24x + 144 = x2 + 21x + 90

3x = -54 ⇒ x = −5/43 = -18

Question 5.

Find the number of terms in the following G.P.

(i),4, 8, 16, …, 8192

(ii) 1/3,1/9,1/27,……1/2187

Solution:

(i) 4, 8, 16, …… 8192

Question 6.

In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.

Solution:

In a G.P

tn = arn-1

t9 = 32805

t6 = 1215

t12 = ?

Let

t9 = ar8 = 32805 ………(1)

t6 = ar5 = 1215 ………. (2)

Question 7.

Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.

Solution:

Here r = 2, t8 = 768

t8 = 768 (tn = arn-1)

a. r8-1 = 768

ar7 = 768 …..(1)

10th term of a G.P. = a.r 10-1

= ar9

= (ar7) × (r2)

= 768 × 22 (from 1)

= 768 × 4 = 3072

∴ 10th term of a G.P. = 3072

Question 8.

If a, b, c are in A.P. then show that 3a,3b,3c are in G.P.

Solution:

If a, b, c are in A.P

t2 – t1 = t3 – t2

b – a = c – b

2b = c + a

To prove that 3a, 3b, 3c are in G.P

⇒ 32b = 3c+a + a [Raising the power both sides]

⇒ 3b.3b = 3c.3a

⇒3b/3a=3c/3b

⇒t2/t1=t3/t1

⇒ Common ratio is same for 3a,3b,3c

⇒ 3a, 3b, 3c forms a G.P

∴ Hence it is proved .

Question 9.

In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2. Find the three terms.

Solution:

Let the three consecutive terms in a G.P are ar, a, a/r.

Their Product = a/r × a × ar = 27

a3 = 27 = 33

a = 3

Sum of the product of terms taken two at a time is 57/2

Question 10.

A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

Solution:

Starting salary = ₹ 60,000

Increase per year = 5%

∴ At the end of 1 year the increase

= 60,0,00 × 5/100

₹ 3000

∴ At the end of first year his salary

= ₹ 60,000 + 3000

I year salary = ₹ 63,000

II Year increase = 63000 × 5/100

At the end of II year, salary

= 63000 + 3150

= ₹ 66150

III Year increase = 66150 × 5/100

= 3307.50

At the end of III year, salary = 66150 + 3307.50

= ₹ 69457.50

IV year increase = 69457.50 × 5/100

= ₹ 3472.87

Question 11.

Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹ 20,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

Offer B: ₹ 22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4th year with respect to the offers A and B?

Solution:

Offer A

Starting salary ₹ 20,000

Annual increase 6%

At the end of

III year ,salary = 22472 + 1348 = 23820

∴ IV year salary = ₹ 23820

Offer B

Starting salary = ₹ 22,000

Salary as per Option A = ₹ 23820

Salary as per Option B = ₹ 24040

∴ Option B is better.

Question 12.

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb-c × yc-a × za-b = 1.

Solution:

a, b, c are three consecutive terms of an AP.

∴ Let a, b, c be a, a + d, a + 2d respectively ………… (1)

x, y, z are three consecutive terms of a GP.

∴ Assume x, y, z as x, x.r, x.r2 respectively ……… (2)

PT : xb-c , yc-a , za-b = 1

Substituting (1) and (2) in LHS, we get

LHS = xa+d-a-2d × (xr)a+2d-a × (xr2)a-a-d

= (x)-d . (xr)2d (xr2)-d

= 1/xd × x2d . r2d × 1/xdr2d = 1 = RHS