# Progressions (AP,GP, HP) | Set-2

**Question 1:** Find the 1st term of the AP whose 7th and 11th terms are respectively 37 and 57. **Solution :**We know nth term of AP is [a + (n-1)d]

So, 7th term = a + 6d

and 11th term = a + 10d

Given a + 6d = 37 ………(1)

a + 10d = 57……….(2)

Subtract (1) form (2)

4d = 20

d = 5

put value of d in (1)

a + 6×5 = 37

a = 7

Hence, 1st term of the AP is **7**.

**Question 2:** A number 21 is divided into three parts which are in AP and sum of their squares is 155. Find the largest number. **Solution :** Let the three consecutive parts of AP are (a-d), a, (a+d).

Given that

(a-d) + a + (a+d) = 21

3a = 21

a = 7

Again, (a-d)^{2} + a^{2} + (a+d)^{2} = 155

a^{2} + d^{2} – 2ad + a^{2} + a^{2} + d^{2} + 2ad = 155

3a^{2} + 2d^{2} = 155

put value of a

3(7)^{2} + 2d^{2} = 155

2d^{2} = 155 – 147

d^{2} = 4

d = ∓2

Hence, the largest part is (a+d) = 7+2 = 9

**Question 3:** How many natural numbers between 200 to 500 are multiples of 3? **Solution :** The series have multiple starting from 201, 204, ……..498

It becomes an AP having first term 210 and common difference 3.

Total number of natural numbers= [(last term – first term)/diff] + 1

= [498 – 201)/3] + 1

= 297/3 + 1

= 100

**Question 4:** The 8th term of a GP is 16 times the 4th term. What will be the first term when its sixth term is 64. **Solution :** We know nth term of GP = ar^{n-1}

8th term = ar^{7} and 4th term =ar^{3}

Given that

ar^{7} = 16ar^{3}

=> r^{4} = 16

=> r = 2

Given ar^{5} = 64

put value of r above

a(2)^{5} = 64

a = 64/32 = 2

Hence, first term of the GP is 2.

**Question 5:** A and B are two numbers whose AM is 61 and GM is 11. What will be the possible value of A? **Solution :** AM is 61 means there sum is 2×61 = 122

and, GM is 11 means there product is 11^{2} = 121

Only possible values for A and B is 121 and 1.

So, value of A is 121.

**Question 6:** Find the number of terms in the series 1/8, 1/2, 2…….8192. **Solution :** 1st term = 1/8

Last term = 8192

Number of terms in GP

ar^{n-1} = 8192

(1/8)(4)^{n-1} = 2^{13}

4^{n-1} = 2^{16}

2^{2n-2} = 2^{16}

2n-2 = 16

n = 9

Hence, number of terms in GP is **9**.

**Question 7:** A rubber ball rebounds (5/6)th of its height after striking to the ground form which it has fallen.Find the total distance that it travels before coming to rest, if it is gently dropped from a height of 360 meters. **Solution :** It becomes an infinite sum of series.

So, use a/(1-r) to calculate the distance

Ball rebounds to 5/6 of its height -> 360x(5/6) = 300

[360/1-(5/6)] + [300/1-(5/6)]

[360/(1/6)] + [300/(1/6)]

= 2160 + 1800

= 3960

Hence, the total distance travelled by the ball is **3960 meters**.

**Question 8:** A man joins a company XYZ in January 2019 and he receive his first salary Rs 1000. After every month he gets an increment of Rs 500. What will be his salary after completion of 5 years of his service. **Solution :** It is an AP 1000, 1500, 2000, ……..so on

In 5 years, there are total 60 months.

We need to find the 60th term of the series.

Common difference d = 500

a_{60} = a + (n-1)d

a_{60} = 1000 + 59x 500

a_{60} = 1000 + 29500 = 30500

After completion of 5 years of service his salary will be **Rs 30500**.