Explore topic-wise InterviewSolutions in .

a) 2, 3, 5, 7, 8, 10, 15, …… is not an A.P. 

∵ a₂ – a₁ = 3 – 2 = 1 

a₃ – a₂ = 5 – 3 = 2 

a₄ – a₃ = 7 – 5 = 2 

i.e., The difference between any two successive terms is not same throughout the series. 

(or) 

Every number is not formed by adding a fixed number to its preceding term.

b) The given list does not form an A.P, since each term is not obtained by adding a fixed number to its preceding term.

c) -1,-3,-5,-7,….. is an A.P. 

a₂ – a₁ = – 3 – (- 1) = -3 + 1 = -2 

a₃ – a₂ = – 5 – (-3) = -5 + 3 = -2 

a₄ – a₃ = – 7 – (- 5) = -7 + 5 = -2 

Every number is formed by adding a fixed number to its preceding term,

Common difference d = a₂ – a₁ 

a) 148 – 147 = 1 

b) -3.0 – (-3.1) = 0.1 

c) 900 – 950 = -50 

d) 250 – 200 = 50

e) 100 – 50 = 50 

Common difference is positive when a₂ > a₁

a = 100; d = 1000 then A.P. is 100, 1100, 2100, 3100, 4100, ……

a = 80, d = -7 then A.P. is 80, 73, 66, 59, 52, ……

Correct answer is

(B) 465

  = n(n+1)/2

   =30(30+1)/2
   =465

Correct answer is

(C) -2, -4, -6, -8

Correct answer is:- (B) is an A.P. Reason d = 4

Explanation :- Let the common difference be d

 ⇒ T₁ = -10, T_{2 =} -6 

⇒ d = T_{2 -} T₁ = -6 + 10

∴ d = 4 

Hence, the sequence is an AP with common difference as 4. 

Correct answer is

(B) is an A.P. Reason d = 4

3, 5, 7, 9, 11, 13, 15.... is an arithmetic progression.

Here the common difference between two consecutive terms is 2.

A sequence in which the difference between any two consecutive terms is a constant is called as arithmetic progression.

here is your answer is - ARITHMETIC  PROGERATION

EXPLANATION- A sequence in which the difference between any two consecutive terms is a constant is called arithmetic progression.

hopefully, it help!

To Find: n^th term of the AP

Given: The series is 8, 3, –2, –7, –12, ….

a₁ = 8, a₂ = 3 and d = 3–8 = –5

(Where a = a₁ is first term, a₂ is second term, an is n^th term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n = a₁ + (n - 1)(–5) = 8 – (5n – 5) = 8 – 5n + 5 = 13 – 5n

So the n^th term of AP is equal to 13 – 5n

Let nth term is 88.

AP is 3, 8, 13, 18, …

Here,

First term = a = 3

Common difference = d = 8 – 3 = 5

nth term of AP is a_n = a + (n – 1) d

Now,

88 = 3 + (n – 1)(5)

88 – 3 = (n – 1) x 5

n – 1 = 88 /5

or n = 17 + 1 = 18

Therefore: 88 is the 18th term.

First two digit number divisible by 3 = 12

Last two digit number divisible by 3 = 99

An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.

A.P = 12,15,18,…,99

here

First term (a) = 12

Common difference (d) = 3

Let us consider there are n numbers then

a_n = 99

a + (n – 1)d = 99

12 + (n – 1)3 = 99

12 + 3n – 3 = 99

n = 29+1

n = 30

Hence there are 30, two digit numbers which are divisible by 3.

S_n \(= \frac{n}{2}[2a + (n-1)d]\)

\(= \frac{n}{2} [10 + (n-1)-3]\)

= \(\frac{n}{2} [13 - 3n]\)

For the given AP the first term a is 1, and common difference d is a difference of the second term and first term, which is 3 - 1 = 2 

To find : The sum of given AP 

The formula for sum of AP is given by,

s = \(\frac{n}{2}\)(2a + (n-1)d)

Substituting the values in the above formula,

s = \(\frac{12}{2}\)(2+(11)(2))

s = 6 × 24 

s = 144

For the given AP the first term a is 50, and common difference d is a difference of second term and first term, which is 46 - 50 = - 4 

To find : The sum of given AP 

The formula for sum of AP is given by

s = \(\frac{n}{2}\)(2a + (n-1)d)

Substituting the values in the above formula,

s = \(\frac{10}{2}\)(100 + (9)(-4))

s = 5 × 64 

s = 320

Given, A.P. 3, 10, 17, ….

Here, a = 3 and d = 10 – 3 = 7

According the question,

a_n = a₁₃ + 84

Using a_n = a + (n – 1)d,

3 + (n – 1)7 = 3 + (13 – 1)7 + 84

3 + 7n – 7 = 3 + 84 + 84

7n = 168 + 7

n = 175/7

n = 25

Therefore, it the 25^th term which is 84 more than its 13^th term.

It is given that (5x + 2),(4x - 1) and (x + 2) are in AP.

∴ (4x - 1) - (5x + 2) = (x + 2) - (4x - 1)

\(\Rightarrow\) 4x - 1 - 5x - 2 = x + 2 - 4x + 1

\(\Rightarrow\) -x - 3 = -3x + 3

\(\Rightarrow\) 3x - x = 3 + 3

\(\Rightarrow\) 2x = 6

\(\Rightarrow\) x = 3

Hence, the value of x is 3.

Correct option is (A) x/3

Given sequence is \(x,\frac{4x}{3},\frac{5x}{3},2x,.......\)

\(\therefore a_1=x,a_2=\frac{4x}{3},a_3=\frac{5x}{3},a_4=2x\)

Now, \(a_2-a_1=\frac{4x}3-x\)

\(=\frac{4x-3x}3=\frac x3,\)

\(a_3-a_2=\frac{5x}3-\frac{4x}3=\frac{x}3\)

\(a_4-a_3=2x-\frac{5x}3\)

\(=\frac{6x-5x}3=\frac x3\)

\(\because\) \(a_4-a_3\) \(=a_3-a_2\) \(=a_2-a_1\) \(=\frac{x}{3}\)

\(\therefore\) Common difference \(=a_4-a_3=a_3-a_2=a_2-a_1=\frac x3\)

Correct option is A) x/3

Given: (5x + 2), (4x – 1) and (x + 2) are terms in AP.

So, d = (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)

2(4x – 1) = (x + 2) + (5x + 2)

8x – 2 = 6x + 2 + 2

8x – 2 = 6x + 4

8x – 6x = 4 + 2

or x = 3

The value of x is 3.

If (a – b)², (a² + b²) and (a + b)² have to be in A.P. then,

It should satisfy the condition,

2b = a + c [for a, b, c are in A.P]

Thus,

2 (a² + b²) = (a – b)² + (a + b)²

2 (a² + b²) = a² + b² – 2ab + a² + b² + 2ab

2 (a² + b²) = 2a² + 2b² = 2 (a² + b²)

LHS = RHS

Hence proved

Let the two A.Ps be A.P₁ and A.P₂

For A.P₁ the first term = a and the common difference = d

And for A.P₂ the first term = b and the common difference = d

So, from the question we have

a₁₀₀ – b₁₀₀ = 100

(a + 99d) – (b + 99d) = 100

a – b = 100

Now, the difference between their 1000^th terms is,

(a + 999d) – (b + 999d) = a – b = 100

Therefore, the difference between their 1000^th terms is also 100.

Given,

x + 1, 3x and 4x + 2 are in A.P.

So, the common difference between the consecutive terms should be the same.

3x – x – 1 = 4x + 2 – 3x

⇒ 2x – 1 = x + 2

⇒ 2x – x = 2 + 1

⇒ x = 3

Therefore, x = 3

Given,

(8x + 4), (6x – 2) and (2x + 7) are in A.P.

So, the common difference between the consecutive terms should be the same.

(6x – 2) – (8x + 4) = (2x + 7) – (6x – 2)

⇒ 6x – 2 – 8x – 4 = 2x + 7 – 6x + 2

⇒ -2x – 6 = -4x + 9

⇒ -2x + 4x = 9 + 6

⇒ 2x = 15

Therefore, x = 15/2

Given terms: x+1, 3x and 4x+2 

Since, the given terms are in A.P. Therefore, their common difference will be same 

3x – (x + 1) = (4x + 2) – 3x 

3x – x - 1 = 4x + 2 – 3x 

2x - 1 = x + 2 

x = 3

We know that, if three numbers are a,b,c are in AP then, 2b = a + c 

According to question, (8x+4), (6x-2), (2x+7) are in AP. 

Hence, 

2 (6x – 2) = 8x + 4 + 2x + 7 

12x – 4 = 10x + 11 

2x = 15 

x = 15/2 = 7.5

Let the numbers be (a – d), a, (a+ d) 

a – d + a + a+ d = 12 

3a = 12 

a = 4

According to question, 

(a – d)³ + a³ + (a + d)³ = 288 

a ³ – d³ – 3ad (a – d) + a³ + a³ + d³ + 3ad (a + d) = 288 

3a³ – 3a²d + 3ad² + 3a²d + 3ad² = 288 

3(4)³ + 6(4) d² = 288 

24d² = 96

d² = 4

d = \(\sqrt{4}\)

Therefore, when a = 4 and d = 2 

a – d = 4 – 2 = 2 

a + d = 4 + 2 = 6 

When a = 4 and d = -2 

a – d = 4 + 2 = 6 

a + d = 4 – 2 = 2

Total/Sum of all cash prizes = Rs. 700 

Each prize differs by Rs. 20 

Let the prizes (in ascending order) be x, x + 20, x + 40, x + 60, x + 80, x + 100, x + 120 

∴ Sum of the prizes = S₇ = \(\frac{n}{2}\)(a + l) 

⇒ 700 = \(\frac{7}{2}\) [x + x + 120] 

⇒ 700 × \(\frac{7}{2}\) = 2x + 120 

⇒ 100 = x + 60 

⇒ x = 100 – 60 = 40 

∴ The prizes are 160, 140, 120, 100, 80, 60, 40.

Correct option is B) \(\cfrac{n(n+1)(n+2)}3\)

Correct option is (D) \(\frac{n(4n^2-1)}{3}\)

\(S_n=1^2+3^2+5^2+......+\) upto n terms

\(=1^2+3^2+5^2+......+(2n-1)^2\)

\(=\sum(2n-1)^2\)

\(=\sum(4n^2-4n+1)\)

\(=4\sum n^2-4\sum n+\sum1\)

\(=\frac{4n(n+1)(2n+1)}6-\frac{4n(n+1)}2+n\)

\(=n\left(\frac23(2n^2+3n+1)-2(n+1)+1\right)\)

\(=\frac n3(4n^2+6n+2-6n-6+3)\)

\(=\frac n3(4n^2-1)\)

Correct option is D) \(\cfrac{n(4n^2-1)}3\)

Correct option is A) \(\cfrac{n(n+1)(n+2)}3\)

Correct option is (C) \(2n^2(n+1)^2\)

\(S_n=2^3+4^3+6^3+.......+\) upto n terms

\(=2^3+4^3+6^3+.......+(2n)^3\)

\(=\sum(2n)^3\) \(=\sum8n^3\)

\(=8\sum n^3\) \(=8\left(\frac{n(n+1)}2\right)^2\)

\(=\frac{8n^2(n+1)^2}4\)

\(=2n^2(n+1)^2\)

Correct option is C) 2n² (n + 1)² 

a_p = q => A + (p - 1) D = q ......(i)

& a_q = p => A + (q - 1) D = p

Solve (i) & (ii) to get D = - 1 & A = p + q - 1

So, a_n = A + (n - 1) D

a_n = (p + q - 1) + (n - 1) (-1) 

a_n = p + q - n.

Correct option is B) \(\cfrac{n(n+1)(2n+1)}6\)

a = 22, d = -4 

S_n = 64 

64 = \(\frac{n}{2}\) (2a + (n – 1) d) 

n [44 – (n – 1)4] = 128 

n (44 – 4n + 4) = 128 

48n – 4n² = 128 

n² – 12n + 32 = 0 

n² – 8n – 4n + 32 = 0 

(n – 8) (n – 4) = 0 

n = 8 or n = 4

Given: S_n = n² + 2n …(i)

S_n-1 = (n – 1)² + 2(n – 1) = n² + 1 – 2n + 2n – 2 = n² - 1 …(ii)

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

t_n = S_n – S_n-1 = n² + 2n – n² + 1 = 2n + 1

The nth term of an AP is 2n + 1.

Given: 4, 8, 16, …… 

where, a₁ = 4; a₂ = 8; a₃ = 16, …… 

\(\frac{a_2}{a_1}\) = 8/4 = 2

\(\frac{a_3}{a_2}\) = 16/8 = 2

∴ r = \(\frac{a_2}{a_1}\)= \(\frac{a_3}{a_2}\) = 2 

Hence 4, 8, 16, … is a G.P. 

where a = 4 and r = 2 

a₄ = a . r³ = 4 × 2³ = 4 × 8 = 32 

a₅ = a . r⁴ = 4 × 2⁴ = 4 × 16 = 64 

a₆ = a . r⁵ = 4 × 2⁵ = 4 × 32 = 128

(i) 0 and 50

a = 1, l = 49, n = 25

S_{25 = \(\cfrac{25}{2}[a + l]\)}

     =_{\(\cfrac{25}{2}[1 + 49]\)}

\(\frac{25}{2} (50) = 625\)

(ii) 100 and 200 a = 101, 

l = 199, n = 50

S₅₀ = _{\(\cfrac{50}{2}[a + l]\)}

      = 25 (101 + 199)

      = 25 (300) = 7500

3-digit numbers: 100, 101,…….,999

3-digit numbers divisible by 9 : 108, 117, 126, 135, …, 999

Here, a = 108, d= 9, l = 999

a_n (l) = a + (n – 1) d

999 = 108 + (n – 1) x 9

(n – 1) x 9 = 999 – 108 = 891

n – 1 = 99

n = 100

3-digit natural numbers: 100, 101,…….. 990 and

3-digit natural numbers divisible by 7: 105, 112, 119, 126, …, 994

Here, a = 105, d= 7, l = 994

a_n = (l) = a + (n – 1) d

994 = 105 + (n – 1) x 7

994 – 105 = (n – 1) 7

n – 1 = 127

n = 128

Answer: There are 128 required numbers.

Given: (2n – 1), (3n + 2) and (6n – 1) are in AP

So, (3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)

(3n + 2) + (3n + 2) = 6n – 1 + 2n – 1

6n + 4 = 8n – 2

8n – 6n = 4 + 2

Or n = 3

Numbers are:

2 x 3 – 1 = 5

3 x 3 + 2 = 11

6 x 3 – 1 = 17

Answer: (5, 11, 17) are required numbers.

Given: a, 9, b, 25 are in AP.

So, 9 – a = b – 9 = 25 – b

b – 9 = 25 – b

b + b = 22 + 9 = 34

or b = 17

And,

a – b = a – 9

9 + 9 = a + b

a + b = 18

a + 17 = 18

or a = 1

Answer: a = 18, b= 17

Here A.P = k, 2k –1, 2k + 1 

Since the numbers are in A.P their common difference (d) should be same 

d=a₂–a₁ = a₃–a₂ 

2k–1–k = 2k + 1– (2k –1) 

k– 1 = 2 

k = 3

Given: 18, a, (b – 3) are in AP

a – 18 = b – 3 – a

a + a – b = -3 + 18

2a – b = 15. Answer!

Correct option is (A) 0

Given that \(p^{th},q^{th}\;\&\;r^{th}\) terms of a G.P. are a, b and c.

\(i.e.,a_p=a,a_q=b\;\&\;a_r=c\)

Let A and R be first term and common difference of given G.P.

\(\therefore\) \(AR^{p-1}=a\)       _______________(1)

\(AR^{q-1}=b\)           _______________(2)

and \(AR^{r-1}=c\)    _______________(3)

Now, \(\sum(q-r)\,log\,a\) \(=(q-r)\,log\,a+(r-p)\,log\,b+(p-q)\,log\,c\)

\(=log\,a^{q-r}+log\,b^{r-p}+log\,c^{p-q}\)

\(=log\,(a^{q-r}.b^{r-p}.c^{p-q})\)

\(=log\,\left((AR^{p-1})^{q-r}.(AR^{q-1})^{r-p}.(AR^{r-1})^{p-q}\right)\)     (From (1), (2) & (3))

\(=log\,(A^{(q-r)+(r-p)+(p-q)}\,R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)})\)

\(=log\,(A^{q-q-r+r-p+p}\,R^{(pq-pr-q+r)+(qr-qp-r+p)+(rp-rq-p+q)})\)

\(=log\,(A^{0}\,R^{pq-pq-pr+pr-q+q+r-r+qr-qr+p-p})\)

\(=log\,(A^{0}\,R^{0})\)

\(=log\,1=0\)

Hence, \(\sum(q-r)\,log\,a=0\)

Correct option is A) 0

It is given that k, (2k - 1) and (2k + 1) are the three successive terms of an AP.

∴ (2k - 1) - k = (2k + 1) - (2k - 1)

\(\Rightarrow\) k - 1 = 2

\(\Rightarrow\) k = 3

Hence, the value of k is 3.

Given: k, (2k – 1) and (2k + 1) are the three successive terms of an AP.

So, (2k – 1) – k = (2k + 1) – (2k – 1)

2 (2k – 1) = 2k + 1 + k

4k – 2 = 3k + 1

4k – 3k = 1 + 2

or k = 3

The value of k is 3.

Let a be the first term and d be the common difference.

Now, a3=-13

=> a + 2d = -13 ---------(1)

now, a6 = 2

=> a + 5d = 2 -----------(2)

Subtracting (1) from (2) we get

5d-2d = 2+13

3d = 15

d = 5

Now, from equation (1) we get

a = -13 - 2x5 = -23

The required A.P is given by

a, a+d, a+2d, ...........

So, A.P is -23, -18, -13, -8, -3, 2,...... 

a) A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. n^th term (or general term) of sequence is denoted by a(n) or a_n

b) Given that squares of terms of the sequence are also terms so the sequence.

i.e. a_n² = a_n

= a_n² - a_n =0

= a_n (a_n -1) = 0

= a_n = 0 or a_n -1 = 0

= a_n = 0 or a_n = 1

Hence, if squares of terms of the sequence are also terms of the sequence than that sequence is either a_n = 0 (zero sequence) or a_n = 1 (Constant sequence whose terms are 1)

Here,

a₁ = 0

a₂ = ¼

a₃ = ½

a₄ = ¾

a₂ – a₁ = ¼ – 0 = ¼

a₃ – a₂ = ½ – ¼ = ¼

a₄ – a₃ = ¾ – ½ = ¼

Since, difference of successive terms are equal,

Hence, 0, 1/4, 1/2, 3/4… is an AP with common difference ¼.

Therefore, the next three term will be,

¾ + ¼ , ¾ + 2(¼), ¾ + 3(¼)

1, 5/4 , 3/2

(i) We have a₂ – a₁ = 10 – 4 = 6

a₃ – a₂ = 16 – 10 = 6

a₄ – a₃ = 22 – 16 = 6

i.e., a_{k + 1} – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = 6.

The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

(ii) a₂ – a₁ = – 1 – 1 = – 2

a₃-a₂=-3-(-1)=-3+1=-2

a₄-a₃= -5-(-3)=-5+3=-2

i.e., a_{k + 1} – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = – 2.

The next two terms are:

– 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9

(iii) a₂ – a₁ = 2 – (– 2) = 2 + 2 = 4

a₃ – a₂ = – 2 – 2 = – 4

As a₂-a₁   a₃-a₂,  the given list of numbers does not form an AP.

(iv) a₂ – a₁ = 1 – 1 = 0

a₃ – a₂ = 1 – 1 = 0

a₄-a₃=2-1=1

3 = 2 – 1 = 1

Here, a₂ – a₁ = a₃ – a₂ ≠a₄ – a₃.

So, the given list of numbers does not form an AP.

Here, a =3/2, d=1/2-3/2=-1.

Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP.