Sequence And Series

Given first term is ‘a’ and sum of first p terms is S_p = 0 we have to find the sum of next q terms

Total terms are p + q

So, sum of all terms minus the sum of the first p terms will give the sum of next q terms

But sum of first p terms is 0 hence sum of next q terms will be the same as sum of all terms

So, we have to find sum of p + q terms

Sum of n terms of AP is given by

Where a is first term and d is the common difference

Using the given hint

⇒ Required sum = S_p+q – S_p

Using S_n formula

Now we have to find d

We use the given S_p = 0 to find d

⇒ 0 = 2a + (p – 1) d

Put this value of d in (i)

⇒required sum

Hence proved.

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 200rs 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

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

Given that S₂₀ = 66000

⇒ 66000 = 10(2a + 19 × 200)

⇒ 6600 = 2a + 3800

⇒ 6600 – 3800 = 2a

⇒ 2a = 2800

⇒ a = 1400

Hence amount saved in first year is 1400 Rs.

His 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) 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

Where a is the first term and d is common difference

⇒ S₁₂ = 6(10400 + 11(320))

⇒ S₁₂ = 6(10400 + 3520)

⇒ S₁₂ = 6(13920)

⇒ S₁₂ = 83520

Hence his total earnings in first year is 83520rs

The n^th term of GP is given by t_n = ar^n-1 where a is the first term and r is the common difference

p^th term is given as q

⇒ t_p = ar^p-1

⇒ q = ar^p-1

q^th term is given as p

⇒ t_q = ar^q-1

⇒ p = ar^q-1

Using (a) and (b)

(p + q)^th term is given by

⇒ t_p+q = ar^p+q-1

⇒ t_p+q = (ar^p-1)r^q

But t_p = ar^p-1 and the p^th term is q

⇒ t_p+q = qr^q

But

Hence proved the (p + q)^th term is .

The 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

Where a is the first term and d is common difference

⇒ 384 = 10n + 2n² – 2n

⇒ 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.

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

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 = 1/2

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

Hence the side of 6^th equilateral triangle is cm and hence its perimeter would be thrice its side length because its an equilateral triangle

Perimeter of 6^th equilateral triangle inscribed is

At start he has to run 24m to get the first potatoe then 28 m as the next potatoe 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 + …upto 20 terms

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

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

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

That is

There are 20 terms hence n = 20

⇒ S₂₀ = 10(48 + 19(4))

⇒ S₂₀ = 10(48 + 76)

⇒ S₂₀ = 10 × 124

⇒ S₂₀ = 1240 m

Using (a)

⇒ 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

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 where

A is first term and d is common difference

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

⇒ 8000 = 8(2a – 15d)

⇒ 1000 = 2a – 15d …(i)

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 …(ii)

Subtract (ii) from (i) that is (i) – (ii)

⇒ 1000 – 275 = 2a – 15d – a + 15d

⇒ 725 = a

Hence amount received by first place team is 725 Rs.

To prove that:

Multiplying the first term by , the second term by and so on that is rationalizing each term

Using (a + b)(a – b) = a² – b²

As a₁, a₂, a₃,…,a_n are in AP let its common difference be d

a₂ – a₁ = d, a₃ – a₂ = d … a_n – a_n-1 = d

Hence multiplying by -1

a₁ – a₂ = -d, a₂ – a₃ = -d … a_n – a_n-1 = -d

Put these values in LHS

Multiply divide by

Using (a + b)(a – b) = a² – b²

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

Where the t_n = a_n is the last n^th term and a = a₁ is the first term

Hence a_n = a₁ + (n – 1)d

⇒ a₁ – a_n = -(n – 1)d

Substitute a₁ – a_n in LHS

⇒ LHS = RHS

Hence proved

Let the series be S = (3³ – 2³) + (5³ – 4³) + (7³ – 6³) + ...

i) Generalizing the series in terms of i

Using a³ – b³ = (a – b)(a² + ab + b²)

We know that and

⇒ 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 upto n terms is 4n³ + 9n² + 6n

ii) Sum of first 10 terms or upto 10 terms

To find sum upto 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 upto 10 terms is 4960

Sum of first n terms be S_n given as

S_n = 2n + 3n²

We have to find the r^th term that is a_r

Using the hint: the n^th term is given as a_n = S_n – S_n-1

⇒ a_r = S_r – S_r-1

Using S_n = 2n + 3n²

⇒ a_r = 2r + 3r² – (2(r – 1) + 3(r – 1)²)

⇒ a_r = 2r + 3r² – (2r – 2 + 3(r² – 2r + 1))

⇒ a_r = 2r + 3r² – (2r – 2 + 3r² – 6r + 3)

⇒ a_r = 6r – 1

Hence the r^th term is 6r – 1

Let the two numbers be ‘a’ and ‘b’

The arithmetic mean is given by and the geometric mean is given by

We have to insert two geometric means between a and b

Now that we have the terms a, G₁, G₂, b

G₁ will be the geometric mean of a and G₂ and G₂ will be the geometric mean of G₁ and b

Hence and

Square

⇒ G₁² = aG₂

Put

Square both sides

⇒ G₁⁴ = a²(G₁b)

⇒ G₁³ = a²b

Put value of G₁ in

Now we have to prove that

Consider RHS

Substitute values of G₁ and G₂ from (i) and (ii)

⇒ RHS = a + b

Divide and multiply by 2

But

Hence

⇒ RHS = 2A

Hence RHS = LHS

Hence proved

we have to prove that

sec θ₁ sec θ₂ + sec θ₂ sec θ₃ + ... + sec θ_n–1 sec θ_n =

⇒ sind(secθ₁ secθ₂ + secθ₂ secθ₃ + ... + sec θ_n–1 sec θ_n) = tan θ_n – tan θ₁

Consider LHS

Now we have to find value of d in terms of θ so that further simplification can be made

As θ₁, θ₂, θ₃, ..., θ_n are in AP having common difference as d

Hence

θ₂ – θ₁ = d, θ₃ – θ₂ = d, …, θ_n – θ_n-1 = d

Take sin on both sides

sin(θ₂ – θ₁) = sind, sin(θ₃ – θ₂) = sind, …, sin(θ_n – θ_n-1) = sind

Substitute appropriate value of sind for each term in LHS

We know that sin(a – b) = sinacosb – cosasinb

= tan θ₂ – tan θ₁ + tan θ₃ – tan θ₂ + … + tan θ_n – tan θ_n-1

= - tan θ₁ + tan θ_n

= tan θ_n - tan θ₁

⇒ LHS = RHS

Hence proved

The sum of n terms of an AP is given by

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

Given that S_p = q and S_q = p

Subtract (i) from (ii) that is (ii) – (i)

Using a² – b² = (a + b)(a – b)

We have to show that S_p+q = -(p + q)

Using (m) and (n)

⇒ S_p+q = S_p + S_q + pqd

= q + p + pqd

Substitute d from (iii)

= (p + q) – 2(p + q)

= -(p + q)

Now we have to find sum of p – q terms that is S_p-q

Using (m) and (n)

Substitute d from (iii)

Hence sum of p – q terms is

Let the first term of AP be m and common difference as d

Let the GP first term as l and common ratio as s

The n^th term of an AP is given as t_n = a + (n – 1)d where a is the first term and d is the common difference

The n^th term of a GP is given by t_n = ar^n-1 where a is the first term and r is the common ratio

The p^th term (t_p) of both AP and GP is a

For AP

⇒ t_p = m + (p – 1)d

⇒ a = m + (p – 1)d …(u)

For GP

⇒ t_p = ls^p-1

⇒ a = ls^p-1 …(v)

The q^th term (t_q) of both AP and GP is b

For AP

⇒ t_q = m + (q – 1)d

⇒ b = m + (q – 1)d …(w)

For GP

⇒ t_q = ls^q-1

⇒ b = ls^q-1 …(x)

The r^th term (t_r) of both AP and GP is c

For AP

⇒ t_r = m + (r – 1)d

⇒ c = m + (r – 1)d …(y)

For GP

⇒ t_r = ls^r-1

⇒ c = ls^r-1 …(z)

Let us find b – c, c – a and a – b

Using (w) and (y)

⇒ b – c = (q – r)d …(i)

Using (y) and (u)

⇒ c – a = (r – p)d …(ii)

Using (u) and (w)

⇒ a – b = (p – q)d …(iii)

We have to prove that a^b–c.b^c–a.c^a–b = 1

LHS = a^b–c.b^c–a.c^a–b)

Using (v), (x) and (z)

⇒ LHS = (ls^p-1)^b-c.(ls^q-1)^c-a.(ls^r-1)^a-b

= s^p(b-c).s^q(c-a).s^r(a-b)

Substituting values of a – b, c – a and b – c from (iii), (ii) and (i)

= s^p(q-r)d.s^q(r-p)d.s^r(p-q)d

= s^pqd-prd.s^qrd-pqd.s^prd-qrd

= s^{pqd-prd+qrd-pqd+prd-qrd}

= s⁰ = 1

⇒ LHS = RHS

Hence proved

Given: S_n = 3n + 2n²

To find: Common Difference of A.P i.e.‘d’

Consider,

S_n = 3n + 2n² [given]

Putting n = 1, we get

S₁ = 3(1) + 2(1)²

= 3 + 2

S₁ = 5

Putting n = 2, we get

S₂ = 3(2) + 2(2)²

= 6 + 2(4)

= 6 + 8

S₂ = 14

Now, we know that,

S₁ = a₁

⇒ a₁ = 5

and a₂ = S₂ – S₁

= 14 – 5

= 9

∴ Common Difference, d = a₂ – a₁

= 9 – 5

= 4

Hence, the correct option is (d)

Given: Third term of G.P, T₃ = 4

To find: Product of first five terms

We know that,

T_n = ar^{n – 1}

It is given that, T₃ = 4

⇒ ar^{3 – 1} = 4

⇒ ar² = 4 …(i)

Product of first 5 terms = a × ar × ar² × ar³ × ar⁴

= a⁵r^1+2+3+4

= a⁵r¹⁰

= (ar²)⁵

= (4)⁵ [from (i)]

Hence, the correct option is (c)

We know that,

a_n = a + (n – 1)d

So, a₉ = a + (9 – 1)d

⇒ a₉ = a + 8d

and a₁₃ = a + 12d

According to the question,

9 times the 9^th term i.e. a₉ = 13 times the 13^th term i.e. a₁₃

⇒ 9 × a₉ = 13 × a₁₃

⇒ 9(a + 8d) = 13(a + 12d)

⇒ 9a + 72d = 13a + 156d

⇒ 9a – 13a = 156d – 72d

⇒ -4a = 84d

⇒ a = -21d …(i)

Now, we have to find the 22^nd term

∴ a₂₂ = a + 21d

⇒ a₂₂ = -21d + 21d [from (i)]

⇒ a₂₂ = 0

Hence, the correct option is (a)

It is given that x, 2y, 3z are in A.P

∴ 2y – x = 3z – 2y

⇒ 2y + 2y = x + 3z

⇒ 4y = x + 3z

⇒ x = 4y – 3z …(i)

and it is also given that x, y, z are in G.P

∴ Common ratio …(ii)

∴ y × y = x × z

⇒ y² = xz …(iii)

Putting the value of x = 4y – 3z in eq. (iii), we get

y² = (4y – 3z)(z)

⇒ y² = 4yz – 3z²

⇒ 3z² – 4yz + y² = 0

⇒ 3z² – 3yz – yz + y² = 0

⇒ 3z(z – y) – y(z – y) = 0

⇒ (3z – y)(z – y) = 0

⇒ 3z – y = 0 & z – y = 0

⇒ 3z = y & z = y but z and y are distinct numbers

& z ≠ y

[from eq. (ii)]

Hence, the correct option is (b)

The given series is A.P whose first term is ‘a’ and common difference is ‘d’.

We know that,

[∵ S_n = qn²]

⇒ 2qn = 2a + (n – 1)d

⇒ 2qn – (n – 1)d = 2a …(i)

and

[∵ S_m = qm²]

⇒ 2qm = 2a + (m – 1)d

⇒ 2qm – (m – 1)d = 2a …(ii)

Solving eq. (i) and (ii), we get

2qn – (n – 1)d = 2qm – (m – 1)d

⇒ 2qn – 2qm = (n – 1)d – (m – 1)d

⇒ 2q(n – m) = d[n – 1 – (m – 1)]

⇒ 2q(n – m) = d[n – 1 – m + 1]

⇒ 2q(n – m) = d(n – m)

⇒ 2q = d

Putting the value of d in eq. (i), we get

2qn – (n – 1)(2q) = 2a

⇒ 2qn – 2qn + 2q = 2a

⇒ 2q = 2a

⇒ q = a

∴ a = q and d = 2q. So,

⇒ S_q = q² + q²(q – 1)

⇒ S_q = q² + q³ – q²

⇒ S_q = q³

Hence, the correct option is (c)

Given that: S_n denote the sum of first n terms

and S_2n = 3S_n

To find: S_3n : S_n

Now, we know that

⇒ S_2n = n[2a + (2n – 1)d]

As per the given condition of the question, we have

S_2n = 3S_n

⇒ 4an + 2nd(2n – 1) = 6an + 3nd(n – 1)

⇒ 2nd(2n – 1) – 3nd(n – 1) = 6an – 4an

⇒ 4n²d – 2nd – 3n²d + 3nd = 2an

⇒ nd + n²d = 2an

⇒ nd(1 + n) = 2an

⇒ d(n + 1) = 2a …(i)

Now, we have to find S_3n:S_n

So,

[from (i)]

Hence, the correct option is (b)

Here, we have to find the minimum value of 4^x + 4^1–x

So, we use the AM – GM inequality which states that the arithmetic mean of a list of non – negative real numbers is greater than or equal to the geometric mean of the same list .i.e.

⇒ 4^x + 4^1-x ≥ 4

Hence, the correct option is (b)

Given that:

S_n = sum of the cubes of first n natural numbers

s_n = Sum of the first n natural numbers

&

Let T_n be the nth term of the above series

…(i)

We know that,

Sum of cubes of first n natural numbers

and sum of first n natural numbers

∴, eq. (i) becomes

Now, sum of the given series

Hence, the correct option is (a)

Given that: t_n be the nth term of the series

Let S_n = 2 + 3 + 6 + 11 + 18 +…+ t₅₀

Using method of difference, we get

S_n = 2 + 3 + 6 + 11 + 18 + … + t₅₀ …(i)

and S_n = 0 + 2 + 3 + 6 + 11 + … + t₄₉ + t₅₀ …(ii)

Subtracting eq. (ii) from eq. (i), we get

0 = 2 + 1 + 3 + 5 + 7 + … - t₅₀ terms

⇒ t₅₀ = 2 + (1 + 3 + 5 + 7 + … upto 49 terms)

We know that,

⇒ t₅₀ = 2 + 49 + 48 × 49

⇒ t₅₀ = 2 + 49(1 + 48)

⇒ t₅₀ = 2 + 49 × 49

⇒ t₅₀ = 2 + 49²

Hence, the correct option is (d)

Given that:

Volume of a block = 216cm³

and Total Surface area = 252 cm³

Breadth of a rectangular block = a

& Height of a rectangular block = ar

Since, they are in G.P so there common ratio is same.

and we know that,

Volume of a rectangular solid block = L × B × H

[given]

⇒ 216 = a³

⇒ a = 6 …(i)

Now,

Total Surface Area of a block = 2[L×B + B×H + H×L]

[given]

[from (i)]

⇒ 2(r² + r + 1) = 7r

⇒ 2r² + 2r + 2 – 7r = 0

⇒ 2r² – 5r + 2 = 0

⇒ 2r² – 4r – r + 2 = 0

⇒ 2r(r – 2) – 1(r – 2)= 0

⇒ (2r – 1)(r – 2) = 0

⇒ 2r – 1 = 0 & r – 2 = 0

and r = 2

∴ Three unequal edges of the given solid block are:

Case 1: If a = 6 and r = 2, then

Breadth = a = 6

and Height = ar = 6 × 2 = 12

Case 2:

Breadth = a = 6

and

So, the length of the longest side = 12units

Hence, the correct option is (a)

Here, it is given that a, b, c are in G.P

⇒ b = ar …(ii)

Putting the value of b in eq. (i), we get

⇒ c = ar × r

⇒ c = ar²

So,

Ans. For a, b, c to be in G.P. the value of is equal to

Let A.P be a, a + d, a + 2d, …, a + (n – 1)d

Taking first and last term

a₁ + a_n = a + a + (n – 1)d

a₁ + a_n = 2a + (n – 1)d …(i)

Taking second and second last term

a₂ + a_n–2 = (a + d) + [a + (n – 2)d]

= a + d + a + nd – 2d

= 2a + nd – d

= 2a + (n – 1)d

= a₁ + a_n [from (i)]

Taking third term from the beginning and the third from the end

a₃ + a_n-3 = (a + 2d) + [a + (n – 3)d]

= a + 2d + a + nd – 3d

= 2a + nd – d

= 2a + (n – 1)d

= a₁ + a_n [from (i)]

From the above pattern, we see that the sum of terms equidistant from the beginning and end in an A.P. is equal to first term + last term.

Given: Third term of a G.P, T₃ = 4

∴ ar² = 4 …(i)

To find: Product of first five terms

So, First five terms are a, ar, ar², ar³ and ar⁴

∴ Product of first five terms = a × ar × ar² × ar³ × ar⁴

= a⁵r^1+2+3+4

= a⁵r¹⁰

= (ar²)⁵

= (4)⁵ [from (i)]

Ans. The third term of a G.P. is 4, the product of the first five terms is (4)⁵

Let us consider a sequence in G.P

2, 2, 2 …(i)

If two terms are in AP then,

a₂ – a₁ = a₃ – a₂

So, let the sequence (i) is in AP, then

a₂ – a₁ = 2 – 2 = 0

and a₃ – a₂ = 2 – 2 = 0

∴ ar – a = ar² – ar

This means two sequences can be in both A.P and G.P together.

Hence, the given statement is FALSE

Let us consider the progression a, a + d, a + 2d, … and a sequence of prime number 2, 3, 5, 7, …

It is clear that this progression is a sequence but sequence is not a progression because it does not follow a specific pattern.

Hence, the given statement is TRUE

Let us consider an A.P a, a + d, a + 2d, a + 3d

∴ a₂ + a₄ = a + d + a + 3d

= 2a + 4d

= 2(a + 2d)

a₂ + a₄ = 2a₃

[∵a₃ = a + (3 – 1)d = a + 2d]

∴ the third term is equal to half the sum of the second and fourth term.

Hence, the given statement is TRUE

Let us consider two G.P’s

a, ar, ar², ar³, …, ar^n-1

and a₁, a₁r₁, a₁r₁², …, a₁r₁^n–1

Now, we will find the sum of these two G.Ps, we get

(a + a₁) + (ar + a₁r₁) + (ar² + a₁r₁²) + … + (ar^n-1 + a₁r₁^n-1)

Now,

and

Clearly,

So, the sum of two G.Ps is not a G.P because the common ratio is not same.

Now, we will find the difference of the two G.Ps

(a – a₁) + (ar – a₁r₁) + (ar² – a₁r₁²) + … + (ar^n-1 – a₁r₁^n-1)

Now,

and

Clearly,

So, the difference of two G.Ps is not a G.P because the common ratio is not same.

Hence, the given statement is FALSE

Let S_n = an² + bn + c [quadratic expression]

Taking n = 1

∴ S₁ = a(1)² + b(1) + c

= a + b + c

We know that,

S₁ = a₁

∴ a₁ = a + b +c

Taking n = 2

∴ S₂ = a(2)² + b(2) + c

= 4a + 2b + c

So, a₂ = S₂ – S₁

= 4a + 2b + c – (a + b + c)

= 4a + 2b + c – a – b – c

= 3a + b

Taking n = 3

∴ S₃ = a(3)² + b(3) + c

= 9a + 3b + c

So, a₃ = S₃ – S₂

= 9a + 3b + c – (4a + 2b + c)

= 9a + 3b + c – 4a – 2b – c

= 5a + b

Common difference, d = a₂ – a₁

= 3a + b – (a + b +c)

= 3a + b – a – b – c

= 2a – c

and d = a₃ – a₂

= 5a + b – (3a + b)

= 5a + b – 3a – b

= 2a

Here, we see that a₂ – a₁≠ a₃ – a₂

So, it does not represent an AP because common difference is not same.

Hence, the given statement is FALSE

(a) Given:

Here, , and

Hence, the given numbers is in G.P with common ration 1/4

∴ (a) ↔ (iii)

(b) Given: 2, 3, 5, 7

Here, a₂ – a₁ = 3 – 2 = 1

a₃ – a₂ = 5 – 3 = 2

a₄ – a₃ = 7 – 5 = 2

∴ a₂ – a₁≠ a₃ – a₂

Hence, it is not in AP

Now, we will check the ratio

,

So,

So, it is not a G.P

Hence, it is a sequence

∴ (b) ↔ (ii)

(c) 13, 8, 3, -2, -7

Here, a₂ – a₁ = 8 – 13 = -5

a₃ – a₂ = 3 – 8 = -5

a₄ – a₃ = -2 – 3 = -5

∴ a₂ – a₁= a₃ – a₂

Hence, it is an AP

∴ (c) ↔ (i)

Let us assume the required sum = S

Therefore, S = 1² + 2² + 3² + … + n²

Now, we will use the below identity to find the value of S:

n³ - (n – 1)³ = 3n² – 3n + 1

Substituting, n = 1, 2, 3, 4, 5,…, n in the above identity, we get

1³ – (1 – 1)³ = 1³ – 0³ = 3(1)² – 3(1) + 1

2³ – (2 – 1)³ = 2³ – 1³ = 3(2)² – 3(2) + 1

3³ – (3 – 1)³ = 3³ – 2³ = 3(3)² – 3(3) + 1

and so on

n³ - (n – 1)³ = 3n² – 3n + 1

Adding we get,

n³ - 0³ = 3(1² + 2² + 3² + … + n²) - 3(1 + 2 + 3 + 4 + … + n) + (1 + 1 + 1 + 1 + … n times)

Taking common n(n + 1), we get

Thus, the sum of the squares of first n natural numbers =

∴(a) ↔ (iii)

(b) Let us assume the required sum = S

Therefore, S = 1³ + 2³ + 3³ + … + n³

Now, we will use the below identity to find the value of S:

n⁴ - (n – 1)⁴ = 4n³ – 6n² + 4n – 1

Substituting, n = 1, 2, 3, 4, 5,…, n in the above identity, we get

1⁴ – (1 – 1)⁴ = 1⁴ – 0⁴ = 4(1)³ – 6(1)² + 4(1) + 1

2⁴ – (2 – 1)⁴ = 2⁴ – 1⁴ = 4(2)³ – 6(2)² + 4(2) + 1

3⁴ – (3 – 1)⁴ = 3⁴ – 2⁴ = 4(3)³ – 6(3)² + 4(3) + 1

and so on

n⁴ - (n – 1)⁴ = 4n³ – 6n² + 4n – 1

Adding we get,

n⁴ – 0⁴ = 4(1³ + 2³ + 3³ + … + n³) – 6(1² + 2² + 3² + … + n²) +4(1 + 2 + 3 + 4 + … + n) + (1 + 1 + 1 + 1 + … n times)

⇒ n⁴ = 4S – n(n + 1)(2n + 1) + 2n (n + 1) – n

⇒ 4S = n⁴ + n + n(n + 1)(2n + 1) – 2n(n + 1)

⇒ 4S = n(n³ + 1) + n(n + 1)[(2n + 1) – 2]

⇒ 4S = n[n³ + 1 + (n + 1)(2n – 1)]

⇒ 4S = n[n³ + 1 + 2n² – n + 2n – 1]

⇒ 4S = n[n³ + 2n² + n]

⇒ 4S = n²(n² + 2n + 1)

⇒ 4S = n²(n + 1)²

Thus, the sum of the cubes of first n natural numbers =

∴(b) ↔ (i)

(c) Let S_n = 2 + 4 + 6 + … + 2n

= 2(1 + 2 + 3 + … + n)

= n(n + 1)

∴ (a) ↔ (ii)

(d) Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + … + n

Clearly, it is an Arithmetic Progression whose first term = 1, last term = n and number of terms = n.

Using the formula,

Therefore,

or we can say that,

∴ (d) ↔ (iv)