### __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