NCERT Class 11 Maths Chapter 9 – Sequences and Series Solutions

Given the nth term, an = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.

 

Given the nth term, an = 2n

On substituting n = 1, 2, 3, 4, 5, we get

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

Hence, the required terms are 2, 4, 8, 16, and 32.

 

Given,

The nth term of the sequence is an = n2/2n

Now, on substituting n = 7, we get

a7 = 72/27 = 49/ 128

 

Given,

nth term of a sequence an = n (n + 2) 

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a1 = 1(1 + 2) = 3

a2 = 2(2 + 2) = 8

a3 = 3(3 + 2) = 15

a4 = 4(4 + 2) = 24

a5 = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

 

Given the nth term, an = (2n – 3)/6

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..

 

Given the nth term, an = (-1)n-1 5n+1

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 25, –125, 625, –3125, and 15625.

 

On substituting n = 1, 2, 3, 4, 5, we get the first 5 terms.

Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.

 

Given,

The nth term of the sequence is an = 4n – 3

On substituting n = 17, we get

a17 = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a24 = 4(24) – 3 = 96 – 3 = 93

 

Given,

The nth term of the sequence is an = (-1)n-1 n3

On substituting n = 9, we get

a9 = (-1)9-1 (9)3 = 1 x 729 = 729

 

 

On substituting n = 20, we get

 

Given, an = 3an-1 + 2 and a1 = 3

Then,

a2 = 3a1 + 2 = 3(3) + 2 = 11

a3 = 3a2 + 2 = 3(11) + 2 = 35

a4 = 3a3 + 2 = 3(35) + 2 = 107

a5 = 3a4 + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

 

Given,

an = an-1/n and a1 = -1

Then,

a2 = a1/2 = -1/2

a3 = a2/3 = -1/6

a4 = a3/4 = -1/24

a5 = a4/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

 

Given,

a1 = a2, an = an-1 – 1

Then,

a3 = a2 – 1 = 2 – 1 = 1

a4 = a3 – 1 = 1 – 1 = 0

a5 = a4 – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

 

Given,

1 = a1 = a2

an = an – 1 + an – 2, n > 2

So,

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

Thus,

 

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where the first term, a = 1

The common difference, d = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

Sn = n/2 [2a + (n-1)d]

Therefore, the sum of odd numbers from 1 to 2001 is 1002001.

 

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where the first term, a = 105

The common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

Sn = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

 

Given,

The first term (a) of an A.P = 2

Let’s assume d is the common difference of the A.P.

So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20th term of the A.P. is –112.

 

Let’s consider the sum of n terms of the given A.P. as –25.

We known that,

Sn = n/2 [2a + (n-1)d]

where n = number of terms, a = first term, and d = common difference

So here, a = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

 

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, the sum of a certain number of terms of the A.P. is 116.

The number of terms is n.

So, we have

Sn = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n2

3n2 – 53n + 232 = 0

3n2 – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, the 8th term is the last term of the A.P.

a8 = 25 + (8 – 1)(-3)

= 25 – 21

= 4

 

Given, the kth term of the A.P. is 5k + 1.

kth term = ak = a + (k – 1)d

And,

a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

On comparing the coefficient of k, we get d = 5

a – d = 1

a – 5 = 1

⇒ a = 6

 

 

We know that,

Sn = n/2 [2a + (n-1)d]

From the question, we have

On comparing the coefficients of n2 on both sides, we get

d/2 = q

Hence, d = 2q

Therefore, the common difference of the A.P. is 2q.

 

Let a1, a2, and d1, d2 be the first terms and the common difference of the first and second arithmetic progression, respectively.

Then, from the question, we have

Let’s take a and d to be the first term and the common difference of the A.P., respectively.

Then, it is given that

Therefore, the sum of (p + q) terms of the A.P. is 0.

 

Let a1 and d be the first term and the common difference of the A.P., respectively.

Then, according to the question, we have

Now, subtracting (2) from (1), we get

Let’s consider that a and b are the first term and the common difference of the A.P., respectively.

Then, from the question, we have

Hence, the given result is proved.

 

Let’s consider a and b to be the first term and the common difference of the A.P., respectively.

am = a + (m – 1)d = 164 … (1)

The sum of the terms is given by,

Sn = n/2 [2a + (n-1)d]

Let’s assume A1, A2, A3, A4, and A5 to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.

Here, we have,

a = 8, b = 26, n = 7

So,

26 = 8 + (7 – 1) d

6d = 26 – 8 = 18

d = 3

Now,

A1 = a + d = 8 + 3 = 11

A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

 

The A.M between a and b is given by (a + b)/2

Then, according to the question,

Thus, the value of n is 1.

 

Let’s consider a1, a2, … am be m numbers such that 1, a1, a2, … am, 31 is an A.P.

And here,

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = 30/ (m + 1) ……. (1)

Now,

a1 = a + d

a2 = a + 2d

a3 = a + 3d …

Hence, a7 = a + 7d

am–1 = a + (m – 1) d

According to the question, we have

Therefore, the value of m is 14.

 

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105, and so on as the instalment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And then, A.P. is 100, 105, 110, …

Where the first term, a = 100

Common difference, d = 5

So, the 30th term in this A.P. will be

A30  = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30th instalment will be Rs 245.

 

It’s understood from the question that the angles of the polygon will form an A.P. with a common difference d = 5° and first term a = 120°.

And we know that the sum of all angles of a polygon with n sides is 180° (n – 2).

Thus, we can say

Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.

 

Given G.P. is 5/2, 5/4, 5/8, ………

Here, a = First term = 5/2

r = Common ratio = (5/4)/(5/2) = ½

Thus, the 20th term and nth term

Given,

The common ratio of the G.P., r = 2

And, let a be the first term of the G.P.

Now,

a8 = ar 8–1 = ar7

ar7 = 192

a(2)7 = 192

a(2)7 = (2)6 (3)

 

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then, according to the question, we have

a5 = a r5–1 = a r4 = p … (i)

a8 = a r8–1 = a r7 = q … (ii)

a11 = a r11–1 = a r10 = s … (iii)

Dividing equation (ii) by (i), we get

Let’s consider a to be the first term and r to be the common ratio of the G.P.

Given, a = –3

And we know that,

an = arn–1

So, a4 = ar3 = (–3) r3

a2 = a r1 = (–3) r

Then, from the question, we have

(–3) r3 = [(–3) r]2

⇒ –3r3 = 9 r2

⇒ r = –3

a7 = a r 7–1 = a r6 = (–3) (–3)6 = – (3)7 = –2187

Therefore, the seventh term of the G.P. is –2187.

 

(a) The given sequence, 2, 2√2, 4,…

We have,

a = 2 and r = 2√2/2 = √2

Taking the nth term of this sequence as 128, we have

Therefore, the 13th term of the given sequence is 128.

(ii) Given the sequence, √3, 3, 3√3,…

We have,

a = √3 and r = 3/√3 = √3

Taking the nth term of this sequence to be 729, we have

Therefore, the 12th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, …

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the nth term of this sequence to be 1/19683, we have

Therefore, the 9th term of the given sequence is 1/19683.

 

The given numbers are -2/7, x, -7/2

Common ratio = x/(-2/7) = -7x/2

Also, common ratio = (-7/2)/x = -7/2x

Therefore, for x = ± 1, the given numbers will be in G.P.

 

Given G.P., 0.15, 0.015, 0.00015, …

Here, a = 0.15 and r = 0.015/0.15 = 0.1 

 

The given G.P. is √7, √21, 3√7, ….

Here,

a = √7 and

he given G.P. is 1, -a, a2, -a3 ….

Here, the first term = a1 = 1

And the common ratio = r = – a

We know that,

 

Given G.P. is x3, x5, x7, …

Here, we have a = x3 and r = x5/x3 = x2

 

et a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 …… (1)

(a/r) (a) (ar) = 1 …….. (2)

From (2), we have

a3 = 1

Hence, a = 1 [Considering real roots only]

Substituting the value of a in (1), we get

1/r + 1 + r = 39/10

(1 + r + r2)/r = 39/10

10 + 10r + 10r2 = 39r

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(5r – 2) (2r – 5) = 0

Thus,

r = 2/5 or 5/2

Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

 

Given G.P. is 3, 32, 33, …

Let’s consider that n terms of this G.P. be required to obtain the sum 120.

We know that,

Here, a = 3 and r = 3

Equating the exponents, we get n = 4

Therefore, four terms of the given G.P. are required to obtain the sum 120.

 

Let’s assume the G.P. to be a, ar, ar2, ar3, …

Then, according to the question, we have

a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128

a (1 + r + r2) = 16 … (1) and,

ar3(1 + r + r2) = 128 … (2)

Dividing equation (2) by (1), we get

r3 = 8

r = 2

Now, using r = 2 in (1), we get

a (1 + 2 + 4) = 16

a (7) = 16

a = 16/7

Now, the sum of terms is given as

 

Given,

a = 729 and a7 = 64

Let r be the common ratio of the G.P.

Then, we know that, an = a rn–1

a7 = ar7–1 = (729)r6

⇒ 64 = 729 r6

r6 = 64/729

r6 = (2/3)6

r = 2/3

And we know that

 

Consider a to be the first term and r to be the common ratio of the G.P.

Given, S2 = -4

Then, from the question, we have

And,

a5 = 4 x a3

ar4 = 4ar2

r2 = 4

r = ± 2

Using the value of r in (1), we have

Therefore, the required G.P is

-4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……

 

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a4 = a r3 = x … (1)

a10 = a r9 = y … (2)

a16 = a r15 = z … (3)

On dividing (2) by (1), we get

 

Given sequence: 8, 88, 888, 8888…

This sequence is not a G.P.

But, it can be changed to G.P. by writing the terms as

Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½

= 64[4 + 2 + 1 + ½ + 1/22]

Now, it’s seen that

4, 2, 1, ½, 1/22 is a G.P.

With the first term, a = 4

Common ratio, r =1/2

We know,

Therefore, the required sum = 64(31/4) = (16)(31) = 496

 

To be proved: The sequence, aA, arAR, ar2AR2, …arn–1ARn–1, forms a G.P.

Now, we have

Therefore, the above sequence forms a G.P., and the common ratio is rR.

 

Consider a to be the first term and r to be the common ratio of the G.P.

Then,

a1 = a, a2 = ar, a3 = ar2, a4 = ar3

From the question, we have

a3 = a1 + 9

ar2 = a + 9 … (i)

a2 = a4 + 18

ar = ar3 + 18 … (ii)

So, from (1) and (2), we get

a(r2 – 1) = 9 … (iii)

ar (1– r2) = 18 … (iv)

Now, dividing (4) by (3), we get

-r = 2

r = -2

On substituting the value of r in (i), we get

4a = a + 9

3a = 9

∴ a = 3

Therefore, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3

i.e., 3¸–6, 12, and –24.

 

Let’s take A to be the first term and R to be the common ratio of the G.P.

Then, according to the question, we have

ARp–1 = a

ARq–1 = b

ARr–1 = c

Then,

aq–r br–p cp–q

= Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r–p) × Ap–q × R(r –1)(p–q)

= Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)

= A0 × R0

= 1

Hence proved.

 

Given the first term of the G.P is a, and the last term is b.

Thus,

The G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.

Then,

b = arn–1  … (1)

P = Product of n terms

= (a) (ar) (ar2) … (arn–1)

= (a × a ×…a) (r × r2 × …rn–1)

= an r 1 + 2 +…(n–1)  … (2)

Here, 1, 2, …(n – 1) is an A.P.

And, the product of n terms P is given by,

 

Let a be the first term and r be the common ratio of the G.P.

Since there are n terms from (n +1)th to (2n)th term,

Sum of terms from(n + 1)th to (2n)th term

a n +1 = ar n + 1 – 1 = arn

Thus, the required ratio =

Thus, the ratio of the sum of the first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is .

 

Given a, b, c, d are in G.P.

So, we have

bc = ad  … (1)

b2 = ac  … (2)

c2 = bd  … (3)

Taking the R.H.S., we have

R.H.S.

= (ab + bc + cd)2

= (ab + ad + cd)2  [Using (1)]

= [ab + d (a + c)]2

= a2b2 + 2abd (a + c) + d2 (a + c)2

= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)

= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2  [Using (1) and (2)]

= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2

= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2

[Using (2) and (3) and rearranging terms]

= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

(a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2

 

Let’s assume G1 and G2 to be two numbers between 3 and 81 such that the series 3, G1, G2, 81 forms a G.P.

And let a be the first term and r be the common ratio of the G.P.

Now, we have the 1st term as 3 and the 4th term as 81.

81 = (3) (r)3

r3 = 27

∴ r = 3 (Taking real roots only)

For r = 3,

G1 = ar = (3) (3) = 9

G2 = ar2 = (3) (3)2 = 27

Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.

 

Given,

The nth term of the series as

an = n2 + 2n

Then, the sum of n terms of the series can be expressed as

 

We know that,

The G. M. of a and b is given by √ab.

Then from the question, we have

By squaring both sides, we get

 

Consider the two numbers to be a and b.

Then, G.M. = √ab.

From the question, we have

 

Given that A and G are A.M. and G.M. between two positive numbers.

And, let these two positive numbers be a and b.

Given the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.

Here we have, a = 30 and r = 2

So, a3 = ar2 = (30) (2)2 = 120

Thus, the number of bacteria at the end of 2nd hour will be 120.

And, a5 = ar4 = (30) (2)4 = 480

The number of bacteria at the end of 4th hour will be 480.

an +1 = arn = (30) 2n

Therefore, the number of bacteria at the end of nth hour will be 30(2)n.

 

Given,

The amount deposited in the bank is Rs 500.

At the end of first year, amount = Rs 500(1 + 1/10) = Rs 500 (1.1)

At the end of 2nd year, amount = Rs 500 (1.1) (1.1)

At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on….

Therefore,

The amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)10

 

 

Let’s consider the roots of the quadratic equation to be a and b.

Then, we have

We know that,

A quadratic equation can be formed as,

x2 – x (Sum of roots) + (Product of roots) = 0

x2 – x (a + b) + (ab) = 0

x2 – 16x + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation is x2 – 16x + 25 = 0

 

Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

It’s seen that,

nth term, an = n ( n + 1)

Then, the sum of n terms of the series can be expressed as

 

Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

It’s seen that,

nth term, an = n ( n + 1) ( n + 2)

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n

Then, the sum of n terms of the series can be expressed as

 

Given series is 3 ×12 + 5 × 22 + 7 × 32 + …

It’s seen that,

nth term, an = ( 2n + 1) n2 = 2n3 + n2

Then, the sum of n terms of the series can be expressed as

 

Given series is 52 + 62 + 72 + … + 202

It’s seen that,

nth term, an = ( n + 4)2 = n2 + 8n + 16

Then, the sum of n terms of the series can be expressed as

Given series is 3 × 8 + 6 × 11 + 9 × 14 + …

It’s found out that,

an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n2 + 15n

Then, the sum of n terms of the series can be expressed as

 

Given series is 12 + (12 + 22) + (12 + 22 + 32 ) + …

Finding the nth term, we have

an = (12 + 22 + 32 +…….+ n2)

Now, the sum of n terms of the series can be expressed as

 

Given,

an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n

Now, the sum of n terms of the series can be expressed as

 

Given,

The nth term of the series as:

an = (2n – 1)2 = 4n2 – 4n + 1

Then, the sum of n terms of the series can be expressed as

 

Let’s take a and d to be the first term and the common difference of the A.P., respectively.

We know that the kth term of an A. P. is given by

ak = a + (k –1) d

So, am + n = a + (m + n –1) d

And, am – n = a + (m – n –1) d

am = a + (m –1) d

Thus,

am + n + am – n  = a + (m + n –1) d + a + (m – n –1) d

= 2a + (m + n –1 + m – n –1) d

= 2a + (2m – 2) d

= 2a + 2 (m – 1) d

=2 [a + (m – 1) d]

= 2am

Therefore, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term

 

Let’s consider the three numbers in A.P. as a – d, a, and a + d.

Then, from the question we have

(a – d) + (a) + (a + d) = 24 … (i)

3a = 24

∴ a = 8

And,

(a – d) a (a + d) = 440 … (ii)

(8 – d) (8) (8 + d) = 440

(8 – d) (8 + d) = 55

64 – d2 = 55

d2 = 64 – 55 = 9

∴ d = ± 3

Thus,

When d = 3, the numbers are 5, 8, and 11 and

When d = –3, the numbers are 11, 8, and 5.

Therefore, the three numbers are 5, 8, and 11.

 

Let’s take a and d to be the first term and the common difference of the A.P., respectively.

So, we have

 

First, let’s find the numbers between 200 and 400, which are divisible by 7.

The numbers are:

203, 210, 217, … 399

Here, the first term, a = 203

Last term, l = 399 and

Common difference, d = 7

Let’s consider the number of terms of the A.P. to be n.

Hence, an = 399 = a + (n –1) d

399 = 203 + (n –1) 7

7 (n –1) = 196

n –1 = 28

n = 29

Then, the sum of 29 terms of the A.P is given by:

Therefore, the required sum is 8729.

 

First let’s find the integers from 1 to 100, which are divisible by 2.

And they are 2, 4, 6… 100.

Clearly, this forms an A.P. with the first term and common difference both equal to 2.

So, we have

100 = 2 + (n –1) 2

n = 50

Hence, the sum is

Now, the integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This also forms an A.P. with the first term and common difference both equal to 5.

So, we have

100 = 5 + (n –1) 5

5n = 100

n = 20

Hence, the sum is

Lastly, the integers, which are divisible by both 2 and 5, are 10, 20, … 100.

And this also forms an A.P. with the first term and common difference both equal to 10.

So, we have

100 = 10 + (n –1) (10)

100 = 10n

n = 10

Thus, the required sum = 2550 + 1050 – 550 = 3050

Therefore, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

 

Given a, b, c, and d are in G.P.

So, we have

∴b2 = ac … (i)

c2 = bd … (ii)

ad = bc … (iii)

Required to prove (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,

(bn + cn)2 = (an + bn) (cn + dn)

Taking L.H.S.

(bn + cn)2 = b2n + 2bncn + c2n

= (b2)n+ 2bncn + (c2) n

= (ac)n + 2bncn + (bd)n  [Using (i) and (ii)]

= an cn + bncn+ bn cn + bn dn

= an cn + bncn+ an dn + bn dn  [Using (iii)]

= cn (an + bn) + dn (an + bn)

= (an + bn) (cn + dn)

= R.H.S.

Therefore, (an + bn), (bn + cn), and (cn + dn) are in G.P

– Hence proved.

 

We have to first find the two-digit numbers, which, when divided by 4, yield 1 as the remainder.

They are 13, 17, … 97.

As it’s seen that this series forms an A.P. with the first term (a) 13 and common difference (d) 4.

Let n be the number of terms of the A.P.

We know that the nth term of an A.P. is given by

an = a + (n –1) d

So, 97 = 13 + (n –1) (4)

4 (n –1) = 84

n – 1 = 21

n = 22

Now, the sum of n terms of an A.P. is given by,

Therefore, the required sum is 1210.

 

Given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3

Taking x = y = 1 in (1), we have

f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

And, f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81

Thus, f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with the first term and common ratio both equal to 3.

We know that sum of terms in G.P is given by,

And it’s given that,

Hence, the sum of the terms of the function is 120.

Therefore, the value of n is 4.

 

Given that the sum of some terms in a G.P is 315.

Let the number of terms be n.

We know that the sum of terms is

Given that the first term a is 5 and the common ratio r is 2.

Hence, the last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160

Therefore, the last term of the G.P. is 160.

 

Let’s consider a and r to be the first term and the common ratio of the G.P., respectively.

Given, a = 1

a3 = ar2 = r2

a5 = ar4 = r4

Then, from the question we have

r2 + r4 = 90

r4 + r2 – 90 = 0

Therefore, the common ratio of the G.P. is ±3.

 

Let’s consider the three numbers in G.P. to be a, ar, and ar2.

Then from the question, we have

a + ar + ar2 = 56

a (1 + r + r2) = 56

… (1)

Also, given

a – 1, ar – 7, ar2 – 21 form an A.P.

So, (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)

ar – a – 6 = ar2 – ar – 14

ar2 – 2ar + a = 8

ar2 – ar – ar + a = 8

a(r2 + 1 – 2r) = 8

a (r – 1)2 = 8 … (2)

7(r2 – 2r + 1) = 1 + r + r2

7r2 – 14 r + 7 – 1 – r – r2 = 0

6r2 – 15r + 6 = 0

6r2 – 12r – 3r + 6 = 0

6r (r – 2) – 3 (r – 2) = 0

(6r – 3) (r – 2) = 0

r = 2, 1/2

When r = 2, a = 8

When r = ½, a = 32

Thus,

When r = 2, the three numbers in G.P. are 8, 16, and 32.

When r = 1/2, the three numbers in G.P. are 32, 16, and 8.

Therefore, in either case, the required three numbers are 8, 16, and 32.

 

Let’s consider the terms in the G.P.to be T1, T2, T3, T4, … T2n.

The number of terms = 2n

Then, from the question we have

T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1]

T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0

T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1] …… (1)

Now, let the terms in G.P. be a, ar, ar2, ar3, …

Then (1) becomes,

[Using sum of terms in G.P.]

ar = 4a

r = 4

Thus, the common ratio of the G.P. is 4.

 

Let’s consider the terms in A.P. to be a, a + d, a + 2d, a + 3d, … a + (n – 2) d, a + (n – 1)d.

From the question, we have

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]

= 4a + (4n – 10) d

Then, according to the given condition,

4a + 6d = 56

4(11) + 6d = 56 [Since a = 11 (given)]

6d = 12

d = 2

Hence, 4a + (4n –10) d = 112

4(11) + (4n – 10)2 = 112

(4n – 10)2 = 68

4n – 10 = 34

4n = 44

n = 11

Therefore, the number of terms of the A.P. is 11.

 

Given,

On cross-multiplying, we have

Also, given

On cross-multiplying, we have

From (1) and (2), we get

b/a = c/b = d/c

Therefore, a, b, c and d are in G.P.

 

Let the terms in G.P. be a, ar, ar2, ar3, … arn – 1…

From the question, we have

Hence, P2 Rn = Sn

 

Let’s assume t and d to be the first term and the common difference of the A.P., respectively.

Then the nth term of the A.P. is given by an = t + (n – 1) d

Thus,

ap = t + (p – 1) d = a  … (1)

aq = t + (q – 1) d = b  … (2)

ar = t + (r – 1) d = c  … (3)

On subtracting equation (2) from (1), we get

(p – 1 – q + 1) d = a – b

(p – q) d = a – b

On subtracting equation (3) from (2), we get

(q – 1 – r + 1) d = b – c

(q – r) d = b – c

Equating both the values of d obtained in (4) and (5), we get

Therefore, the given result is proved.

 

Given a and b are the roots of x2 – 3x + p = 0

So, we have a + b = 3 and ab = p  … (i)

Also, c and d are the roots of x2 – 12x + q = 0

So, c + d = 12 and cd = q  … (ii)

And given a, b, c, d are in G.P.

Let’s take a = x, b = xr, c = xr2, d = xr3

From (i) and (ii), we get

x + xr = 3

x (1 + r) = 3

And,

xr2 + xr3 =12

xr2 (1 + r) = 12

On dividing, we get

When r = 2, x = 3/(1 + 2) = 3/3 = 1

When r = -2, x = 3/(1 – 2) = 3/-1 = -3

Case I:

When r = 2 and x =1,

ab = x2r = 2

cd = x2r5 = 32

Case II:

When r = –2, x = –3,

ab = x2r = –18

cd = x2r5 = – 288

Therefore, in both the cases, we get (q + p): (q – p) = 17:15\

 

Let the two numbers be a and b.

A.M = (a + b)/ 2 and G.M. = √ab

From the question, we have

Given a, b, c are in A.P.

Hence, b – a = c – b  … (1)

And, given that b, c, d are in G.P.

So, c2 = bd  … (2)

Also, 1/c, 1/d, 1/e are in A.P.

So,

Now, required to prove that a, c, e are in G.P. i.e., c2 = ae

From (1), we have

2b = a + c

b = (a + c)/ 2

And from (2), we have

d = c2/ b

On substituting these values in (3), we get

Therefore, a, c, and e are in G.P.

 

(i) Given, 5 + 55 + 555 + …

Let Sn = 5 + 55 + 555 + ….. up to n terms

(ii) Given, .6 + .66 + . 666 + …

Let Sn = 06. + 0.66 + 0.666 + … up to n terms

 

Given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms

∴ nth term = an = 2n × (2n + 2) = 4n2 + 4n

The 20th term,

a20 = 4 (20)2 + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

Therefore, the 20th term of the series is 1680.

 

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an

S = 3 + 7 + 13 + 21 + …. + an – 2 + an – 1 + an

On subtracting both equations, we get

S – S = [3 + (7 + 13 + 21 + 31 + …+ an–1 + an)] – [(3 + 7 + 13 + 21 + 31 + …+ an–1) + an]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – an

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – an

an = 3 + [4 + 6 + 8 + … (n –1) terms]

 

From the question, we have

Therefore, from (1) and (2), we have 9S22 = S3 (1 + 8S1).

nth term of the numerator = n(n + 1)2 = n3 + 2n2 + n

nth term of the denominator = n2(n + 1) = n3 + n2

 

Given, the farmer pays Rs 6000 in cash.

So, the unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

From the question, the interest paid annually will be

12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500

Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)

= 12% of (500 + 1000 + 1500 + … + 6000)

It’s seen that the series 500, 1000, 1500 … 6000 is an A.P. with the first term and common difference both equal to 500.

Let’s take the number of terms of the A.P. to be n.

So, 6000 = 500 + (n – 1) 500

1 + (n – 1) = 12

n = 12

Now,

The sum of the A.P = 12/2 [2(500) + (12 – 1)(500)] = 6 [1000 + 5500] = 6(6500) = 39000

Hence, the total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)

= 12% of 39000 = Rs 4680

Therefore, the tractor will cost the farmer = (Rs 12000 + Rs 4680) = Rs 16680

 

Given, Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

So, the unpaid amount = Rs 22000 – Rs 4000 = Rs 18000

From the question, it’s understood that the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000

Hence, the total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000

= 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of (1000 + 2000 + 3000 + … + 18000)

It’s seen that 1000, 2000, 3000 … 18000 form an A.P. with the first term and common difference both equal to 1000.

Let the number of terms be n.

So, 18000 = 1000 + (n – 1) (1000)

n = 18

Now, the sum of the A.P is given by

Thus,

Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of Rs 171000 = Rs 17100

Therefore, the cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

 

It’s seen that,

The numbers of letters mailed forms a G.P.: 4, 42, … 48

Here, first term = 4 and common ratio = 4

And the number of terms = 8

The sum of n terms of a G.P. is given by

Also, given that the cost to mail one letter is 50 paisa.

Hence, cost of mailing 87380 letters = Rs 87380 x (50/100) = Rs 43690 = Rs 43690

Therefore, the amount spent when the 8th set of the letter is mailed will be Rs 43,690.

 

Given, the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

Hence, the interest in first year = (5/100) x Rs 10000 = Rs 500

So, the amount in the 15th year = Rs

= Rs 10000 + 14 × Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

And the amount after 20 years =

= Rs 10000 + 20 × Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Therefore, the amount in the 15th year is Rs 17,000, and the total amount after 20 years will be Rs 20,000.

 

Given, the cost of the machine = Rs 15625

Also, given that the machine depreciates by 20% every year.

Hence, its value after every year is 80% of the original cost, i.e., 4/5th of the original cost.

Therefore, the value at the end of 5 years =

= 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years will be Rs 5,120.

 

Let’s assume x to be the number of days in which 150 workers finish the work.

Then, from the question, we have

150x = 150 + 146 + 142 + …. (x + 8) terms

The series 150 + 146 + 142 + …. (x + 8) terms is an A.P.

With first term (a) = 150, common difference (d) = –4 and number of terms (n) = (x + 8)

Now, finding the sum of terms

As x cannot be negative [Number of days is always a positive quantity]

x = 17

Hence, the number of days in which the work should have been completed is 17.

But, due to the dropping out of workers, the number of days in which the work is completed

= (17 + 8) = 25