Chapter 9: Sequence and Series

Given the total amount saved in 20 years is 66000 hence S₂₀ = 66000

Let the amount saved in first year be ‘a’

Then as he increases 200 Rs every year then the amount in second year will be ‘a + 200’ then in third year ‘a + 400’ and so on

The sequence will be a, a + 200, a + 400…

The sequence is AP and the common difference is d = 200

There are 20 terms in the sequence as he saved money for 20 years

Using the sum formula for AP S_n = n/2 (2a + (n – 1) d) we get

Where n is number of terms a is the first term and d is the common difference

Given that S₂₀ = 66000

S₂₀ = (20/2) (2a + (20 – 1) 200)

⇒ 66000 = 10(2a + 19 × 200)

⇒ 6600 = 2a + 3800

⇒ 6600 – 3800 = 2a

⇒ 2a = 2800

⇒ a = 1400

Hence amount saved in first year is 1400 Rs.

Given the man’s salary in first month is 5200rs and then it increases every month by 320rs

Hence the sequence of his salary per month will be 5200, 5200 + 320, 5200 + 640…

The sequence is in AP with first term as a = 5200,

Common difference as d = 320

a) Now we have to find salary in 10^th month that is 10^th term of the AP

The n^th term of AP is given by t_n = a + (n – 1) d

Where a is the first term and d is the common difference

We have to find t₁₀

⇒ t₁₀ = 5200 + (10 – 1) (320)

⇒ t₁₀ = 5200 + 2880

⇒ t₁₀ = 8080

Hence salary in 10^th month is 8080rs

b) To get the total earnings in first year we have to add first 12 terms of the sequence that is we have to find S₁₂

The sum of first n terms of AP is given by S_n = n/2 (2a + (n – 1) d)

Where a is the first term and d is common difference

S₁₂ = (12/2) (2 (5200) + (12 – 1) 320)

⇒ S₁₂ = 6(10400 + 11(320))

⇒ S₁₂ = 6(10400 + 3520)

⇒ S₁₂ = 6(13920)

⇒ S₁₂ = 83520

Hence his total earnings in first year is 83520rs

Hence the proof.

Given first day he made 5 frames then two frames more than the previous that is 7 then 9 and so on

Hence the sequence of making frames each day is 5, 7, 9…

The sequence is AP with first term as a = 5 and common difference d = 2

Total number of frames to be made is 192

Let is requires n days hence S_n = 192

The sum of first n terms of AP is given by S_n = n/2 (2a + (n – 1) d)

Where a is the first term and d is common difference

S_n = (n/2) (2(5) + (n – 1) 2)

192 = (n/2) (10 + 2n – 2)

⇒ 384 = 10n + 2n² – 2n

On computing and simplifying we get

⇒ 2n² + 8n – 384 = 0

⇒ n² + 4n – 192 = 0

⇒ n² + 16n – 12n – 192 = 0

⇒ n (n + 16) – 12 (n + 16) = 0

⇒ (n – 12) (n + 16) = 0

⇒ n = 12 and n = -16

But n represents number of days which cannot be negative hence n = 12

Hence number of days required to finish the job is 12 days.

Given the sum of interior angles of a polygon having ‘n’ sides is given by (n – 2) × 180°

Sum of angles with three sides that is n = 3 is (3 – 2) × 180° = 180°

Sum of angles with four sides that is n = 4 is (4 – 2) × 180° = 360°

Sum of angles with five sides that is n = 5 is (5 – 2) × 180° = 540°

Sum of angles with six sides that is n = 6 is (6 – 2) × 180° = 720°

As seen as the number of sides increases by 1 the sum of interior angles increases by 180°

Hence the sequence of sum of angles as number of sides’ increases is 180°, 360°, 540°, 720° …

The sequence is AP with first term as a = 180° and common difference as d = 180°

We have to find sum of angles of polygon with 21 sides

Using (n – 2) × 180°

⇒ Sum of angles of polygon having 21 sides = (21 – 2) × 180°

⇒ Sum of angles of polygon having 21 sides = 19 × 180°

⇒ Sum of angles of polygon having 21 sides = 3420°

Let ABC be the triangle with AB = BC = AC = 20 cm

Let D, E and F be midpoints of AC, CB and AB respectively which are joined to form an equilateral triangle DEF

Now we have to find the length of side of ΔDEF

Consider ΔCDE

CD = CE = 10 cm … D and E are midpoints of AC and CB

Hence ΔCDE is isosceles

⇒ ∠CDE = ∠CED … base angles of isosceles triangle

But ∠DCE = 60° …∠ABC is equilateral

Hence ∠CDE = ∠CED = 60°

Hence ΔCDE is equilateral

Hence DE = 10 cm

Similarly, we can show that GH = 5 cm

Hence the series of sides of equilateral triangle will be 20, 10, 5 …

The series is GP with first term a = 20 and common ratio r = ½

To find the perimeter of 6^th triangle inscribed we first have to find the side of 6^th triangle that is the 6^th term in the series

n^th term in GP is given by t_n = ar^n-1

⇒ t₆ = (20) (1/2)^6-1

⇒ t₆ = 20/ 2⁵

= 20/ (4 × 2³)

⇒ t₆ = 5/8

Hence the side of 6^th equilateral triangle is 5/8 cm and hence its perimeter would be thrice its side length because it’s an equilateral triangle

Perimeter of 6^th equilateral triangle inscribed is 3 × 5/8 = 15/8 cm

Given at start he has to run 24m to get the first potato then 28 m as the next potato is 4m away from first and so on

Hence the sequence of his running will be 24, 28, 32 …

There are 20 terms in sequence as there are 20 potatoes

Hence only to get potatoes from starting point he has to run

24 + 28 + 32 + … up to 20 terms

This is only from starting point to potato but he has to get the potato back to starting point hence the total distance will be twice that is

Total distance ran = 2 × (24 + 28 + 32…) … 1

Let us find the sum using he formula to find sum of n terms of AP

That is S_n = n/2 (2a + (n – 1) d)

There are 20 terms hence n = 20

⇒ S₂₀ = (20/2) (2 (24) + (20 – 1) 4)

On simplification we get

⇒ S₂₀ = 10(48 + 19(4))

⇒ S₂₀ = 10(48 + 76)

On computing we get

⇒ S₂₀ = 10 × 124

⇒ S₂₀ = 1240 m

Using equation 1

⇒ Total distance ran = 2 × 1240

⇒ Total distance ran = 2480 m

Hence total distance he has to run is 2480 m

Let the amount received by first place team be a Rs and d be difference in amount

As the difference is same hence the second-place team will receive a – d and the third place a – 2d and so on

The last team receives 275 Rs

As there are 16 teams and all teams are given prizes hence the sequence will have 16 terms because there are 16 teams

Therefore a, a – d, a – 2d… 275

The sequence is in AP with first term as a and common difference is ‘-d’

As the total prize given is of 8000rs hence

a + a – d + a – 2d … + 275 = 8000

The sum of n terms of AP is given by S_n = n/2 (2a + (n – 1) d) where

A is first term and d is common difference

There are 16 terms n = 16 and the sum S_n = 8000

⇒ 8000 = (16/2) (2a + (16 – 1) (-d))

⇒ 8000 = 8(2a – 15d)

⇒ 1000 = 2a – 15d ……1

The last term of AP is 275 and n^th term of AP is t_n = a + (n – 1) d

The last term is t_n = 275

⇒ 275 = a + (16 – 1) (-d)

⇒ 275 = a – 15d ……2

Subtract 2 from 1 that is 1 – 2

⇒ 1000 – 275 = 2a – 15d – a + 15d

⇒ 725 = a

Hence amount received by first place team is Rs.725

Simplifying we get

⇒ S = 2n (n+1) (2n+1) + 3n (n+1) + n

⇒ S = 2n (2n² + 2n + n + 1) + 3n² + 3n + n

⇒ S = 4n³ + 6n² + 2n + 3n² + 4n

⇒ S = 4n³ + 9n² + 6n

Hence sum up to n terms is 4n³ + 9n² + 6n

ii) Sum of first 10 terms or up to 10 terms

To find sum up to 10 terms put n = 10 in S

⇒ S = 4 (10)³ + 9 (10)² + 6 (10)

⇒ S = 4000 + 900 + 60

⇒ S = 4960

Hence sum of series up to 10 terms is 4960