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551.

The 7th term of an A.P. is – 4 and its 13th term is – 16. Find the A.P.

Answer»

We have

a7 = a + (7 – 1)d = a + 6d = –4 …(1)

and a13 = a + (13 – 1)d = a + 12d = –16 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = –4 – (–16)

⇒ – 6d = –4 + 16

⇒ – 6d = 12

⇒ d = –2

Putting the value of d in eq (1), we get

a + 6(–2) = –4

⇒ a – 12 = –4

⇒ a = 8

Hence, the required AP is 8, 6, 4, 2,…

552.

If the 9th term of an AP is 0, prove that its 29th term is double the 19th term.

Answer»

Prove that: 29th term is double the 19th term (i.e. a29 = 2a19)

Given: a9 = 0

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

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

So a9 = 0 → a + (9 - 1)d = 0

a + 8d = 0

a = (–8d) ….equation (i)

Now a29 = a + (29 - 1)d and a19 = a + (19 - 1)d

a29 = a + 28d and a19 = a + 18d ….equation (ii)

By using equation (i) in equation (ii), we have

a29 = –8d + 28d and a19 = –8d + 18d

a29 = 20d and a19 = 10d

So a29 = 2a19

HENCE PROVED

553.

The nth term of an AP is (3n + 5). Find its common difference.

Answer»

We have

Tn = (3n + 5)

Common difference = T2 - T1

T1 = 3 x 1 + 5 = 8

T2 = 3 x 2 + 5 = 11

d = 11 - 8 - 3

Hence, the common difference is 3.

554.

Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn - k Sn-1 + Sn-2, then k =A. 1 B. 2 C. 3 D. none of these

Answer»

Option : (B)

We know,

Sn = Sn-2+an-1+an 

= Sn-2+a+(n-2)d+a+(n-1)d 

= Sn-2+2a+2nd-3d 

Also, 

Sn-1 = Sn-2+an-1 

= Sn-2+a+(n-2)d 

= Sn-2+a+nd-2d 

Now, 

d = Sn-kSn-1+Sn-2 

= Sn-2+2a+2nd-3d-k(Sn-2+a+nd-2d)+Sn-2 

= (2-k)[Sn-2+a+nd]+d(2k-3) 

Comparing coefficient of both the side we get, 

2-k = 0 and 2k-3 =1 

∴ k = 2

555.

Write one example of finite and infinite A.P. each.

Answer»

Finite A.P.:

Even natural numbers from 4 to 50 :

4, 6, 8, ………… 50.

Infinite A. P.:

Positive multiples of 5 :

5, 10, 15, ………..

556.

If a, b and c are in AP and also in GP, then A) a = b ≠ c B) a ≠ b = c C) a ≠ b ≠ c D) a = b = c

Answer»

Correct option is (D) a = b = c

Given that a, b, c are in AP and also in G.P.

\(\therefore a+c=2b\)     ______________(1)

and \(ac=b^2\)       ______________(2)

\(\Rightarrow ac=(\frac{a+c}2)^2\)              (From (1))

\(\Rightarrow ac=\frac{a^2+c^2+2ac}4\)

\(\Rightarrow a^2+c^2+2ac=4ac\)

\(\Rightarrow a^2+c^2-2ac=0\)

\(\Rightarrow(a-c)^2=0\)

\(\Rightarrow a-c=0\)

\(\Rightarrow a=c\)

Put a = c in equation (1), we get

a+a = 2b

\(\Rightarrow2b=2a\)

\(\Rightarrow b=a\)

Hence, a = b = c

Correct option is D) a = b = c

557.

Find the A.P. whose 1st term is 10 & common difference is 5.

Answer»

Given : First term (a) = 10 & Common difference (d) = 5. 

 A.P. is 10, 15, 20, 25, 30, .....

558.

Find out the common ratio in the GP 2, 2√2 ,4…………..

Answer»

The given GP is 2, 2√2 , 4, ………………

The common ratio = \(\frac{a_2}{a_1}=\frac{2\sqrt{2}}{2}=\sqrt{2}\)

559.

Write the general form of an A.P.

Answer»

If we denote the starting number i.e. the 1st number by ‘a’ and a fixed number to the added is ‘d’ then a, a + d, a + 2d, a + 3d, a + 4d, ...... forms an A.P.

560.

Check whether – 25 is a term in the progression 5, 3, 1, ………….or not ?

Answer»

The given 5, 3, 1,…………..is an arithmetic progression here 

a = 5, d = a2 – a2 = 3 – 5 = -2 

Let -25 is some of ‘n’ th term 

i.e. an – 25 

So an = a + (n-1)d 

-25 = 5 + (n – 1)(- 2)

– 25 – 5 = (n – 1) (-2)

\(\frac{-30}{-2}\) = n – 1 

⇒  n – 1 = 15 and n = 15+ 1 = 16

So – 25 exist at 16th term in above series.

561.

Find the common difference of the following A.P. : 1,4,7,10,13,16 ......

Answer»

4 - 1 = 7 - 4 = 10 - 7 = 13 - 10 = 16 - 13 = 3 (constant). 

Common difference (d) = 3.

562.

Define Arthmetic progression.

Answer»

A sequence is called an A.P., if the difference of a term and the previous term is always same. i.e. d = tn+1tn = Constant for all nN. The constant difference, generally denoted by ‘d’ is called the common difference.

563.

Which term of the A.P. 14, 9, 4, – I, – 6, ... is – 41 ?

Answer»

Here, a = 14, d = 9 – 14 = –5 and an = –41

To find : n

We have,

an = a + (n – 1)d

⇒ –41 = 14 + (n – 1) × (–5)

⇒ –41 = 14 – 5n + 5

⇒ –41 = 19 – 5n

⇒ – 41 – 19 = –5n

⇒ –60 = –5n

⇒ n = 12

Therefore, the 12th term of the given AP is –41.

564.

Which term of A.P. 3, 8, 13, 18, ... is 88?

Answer»

Here, a = 3, d = 8 – 3 = 5 and an = 88

To find : n

We have,

an = a + (n – 1)d

⇒ 88 = 3 + (n – 1) × (5)

⇒ 88 = 3 + 5n – 5

⇒ 88 = –2 + 5n

⇒88 + 2 = 5n

⇒ 90 = 5n

⇒ n = 18

Therefore, the 18th term of the given AP is 88.

565.

Which term of the AP: 3, 8, 13, 18,…, is 78?

Answer»

Given: 3, 8, 13, 18, …… 

Here a = 3; d = a2 – a1 = 8 – 3 = 5 

Let ‘78’ be the nth term of the given A.P.

∴ an = a + (n – 1) d 

⇒ 78 = 3 + (n – 1) 5 

⇒ 78 = 3 + 5n – 5 

⇒ 5n = 78 + 2 

⇒ n = \(\frac{80}{5}\) = 16 

∴ 78 is the 16th term of the given A.P.

566.

Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Answer»

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

Given:

4th term = a4 = 9

Sum of 6th and 13th terms = a6 + a13 = 40

Now,

a4 = a + (4-1)d

9 = a + 3d

a = 9 – 3d ….(1)

And

a6 + a13 = 40

a + 5d + a + 12d = 40

2a +17d = 40

2(9 – 3d) + 17d = 40 (using (1))

d = 2

From (1): a = 9 – 6 = 3

Required AP = 3,5,7,9,…..

567.

Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Answer»

Let a be the first term and d be the common difference of the AP. Then,

a4 = 9

\(\Rightarrow\) a + (4 - 1) d = 9    [an = a + (n - 1)d]

\(\Rightarrow\) a + 3d = 9    ......(1)

Now,

a6 + a13 = 40    (Given)

\(\Rightarrow\) (a + 5d)+(a + 12d) = 40

\(\Rightarrow\) 2a + 17d = 40    ......(2)

From (1) and (2), we get

2(9 - 3d) + 17d = 40

\(\Rightarrow\) 18 - 6d + 17d = 40

\(\Rightarrow\) 11d = 40 - 18 = 22

\(\Rightarrow\) d = 2

Putting d = 2 in (1), we get

a + 3 x 2 = 9

\(\Rightarrow\) a = 9 - 6 = 3

Hence, the AP is 3, 5, 7, 9, 11,…….

The fourth term is given by 

T4=9

a+(n-1)d = 9 where a denotes first term and d denotes common difference 

Here for 4th term n=4

Therefore a+3d=9....(1) × 2

2a +6d =18...(2)

Also T+ T13 = 40

a+(6-1)d + a+(13-1)d = 40

a + 5d + a +12d = 40

2a + 17d = 40....(3)

Solving equations 2 and 3 we get

(3) - (2)

17d -6d = 40-18

11d = 22

d = 2

Now substituting the value of d in equation (2) we have 

2a+6(2) =18

2a= 18-12 = 6

a=3

Therefore the required AP is

3, 5, 7,9,..

568.

If the sum of first ‘n’ terms of an A.P. 2, 5, 8, ……………… is equal to the sum of first n terms of the A.P. 57, 59, 61, ……………… then ‘n’ is equal to ……………..A) 110C) 112B) 111D) 113

Answer»

Correct option is B) 111

569.

Write the expression an - ak for the A.P. a, a + d, a + 2d, ..........Hence, find the common difference of the A.P. for which(i)11th terms is 5 and 13th term is 79.(ii) a10 - a5 = 200(iii) 20th term is 10 more than the 18th term.

Answer»

Let nth term an = a + (n – 1) d

= a + nd – d

kth term, ak = a + (k – 1) d

= a + kd – d

Now,

an - ak = (a + nd – d) – (a + kd – d)

= nd – kd = d (n – k)

(i) a11 = 5

a + 10d = 5 (i)

a13 = 79

a + 12d = 79 (ii)

By subtracting (i) from (ii), we get

2d = 74

d = 37

Hence, the common difference is 37

(ii) a10 = a + 9d (i)

a5 = a + 4d (ii)

a10 – a5 = a + 9d – a – 4d

200 = 5d

d = 40

Hence, common difference is 40

(iii) a20 = a + 19d (i)

a18 = a + 17d (ii)

Given that

a20 = a18 + 10

a + 19d = a + 17d + 10

2d = 10

d = 5

Hence, common difference is 5

570.

The nth term of an AP is (3n + 5). Find its common difference.

Answer»

Nth term = an = 3n + 5 (given)

a(n-1) = 3 (n – 1) + 5 = 3n + 2

Common difference = d = an – a(n-1)

= (3n + 5) – (3n + 2)

= 3n + 5 – 3n – 2

= 3

Therefore, common difference is 3.

571.

If an denotes the nth term of the AP 2, 7, 12, 17, …, find the value of (a30 – a20).

Answer»

Given AP is 2, 7, 12, 17,……..

Here, a = 2, d = 7 – 2 = 5

Now,

an = a + (n – 1) d = 2 + (n – 1) 5 = 5n – 3

a30 = 2 + (30 – 1) 5 = 2 + 145 = 147 and

a20 = 2 + (20 – 1) 5 = 2 + 95 = 97

a30 – a20 = 147 – 97 = 50

572.

Find the nth term of the AP 16, 9, 2, -5, …….

Answer»

AP is 16, 9, 2, -5, ……

Here, first term = a = 16

Common difference = d = 9 – 16 = -7

an = a + (n – 1)d

= 16 + (n – 1) (-7)

= 16 – 7n + 7

= (23 – 7n)

573.

If the nth term of a progression is (4n – 10) show that it is an AP. Find its (i) first term, (ii) common difference (iii) 16 the term.

Answer»

Tn = (4n - 10) [Given]

T1 = (4 x 1 - 10) = -6

T2 = (4 x 2 - 10) = -2

T3 = (4 x3 - 10) = 2

T4 = (4 x 4 - 10) = 6

Clearly, [ - 2 -(-6) ] = [ 2 -(-2) ] = [6 - 2] = 4 (Constant)

So, the terms -6, -2,2,6,......forms an AP.

(i) First term = -6 

(ii) Common difference = 4

(iii) T16 = a + (n - 1)d = a + 15d = -6 + 15 x 4 = 54

574.

If the nth term of a progression is (4n – 10) show that it is an AP. Find its(i) first term,(ii) common difference, and(iii) 16th term.

Answer»

nth term of AP is 4n – 10 (Given)

Putting n = 1, 2, 3, 4, …, we get

At n = 1: 4n – 10 = 4 x 1 – 10 = 4 – 10 = -6

At n = 2: 4n – 10 = 4 x 2 – 10 = 8 – 10 = -2

At n = 3: 4n – 10 = 4 x 3 – 10 = 12 – 10 = 2

At n = 1: 4n – 10 = 4 x 4 – 10 = 16 – 10 = 6

We see that -6, -2, 2, 6,… are in AP

(i) first term = -6

(ii) Common difference = -2 – (-6) = 4

(iii) 16th term:

Using formula: an = a + (n – 1)d

Here n = 16

a16 = -6 + (16 – 1)4 = 54

575.

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Answer»

a10 = 41

a + (10 – 1)d = 41

a + 9d = 41 …(i)

a18 = 73

a + (18 – 1)d = 73

Substituting the value of a from (i),

41 – 9d + 17d = 73

8d = 32

d = 4

a = 41 – 9(4)

a = 41 – 36

a = 5

a26 = a + (26 – 1)d

= 5 + 25(4)

= 5 + 100

= 105

576.

How many two digit numbers are divisible by 4?

Answer»

List of two digit numbers divisible by 4 is 

12, 16, 20, 24, . . . , 96. 

Let’s find how many such numbers are there.

tn = 96, a= 12, d = 4 

From this we will find the value of n. 

tn = 96, By formula, 

96 = 12 +(n - 1) x 4 

 = 12 + 4n - 4 

4n = 88 

n = 22 

There are 22 two digit numbers divisible by 4. 

577.

The 10th and 18th term of an A.P. are 41 and 73 respectively, find 26th term.

Answer»

Given : 

10th term of an A.P is 41, and 

18th terms of an A.P. is 73 

⇒ a10 = 41 and a18 = 73 

We know, 

an = a + (n – 1)d 

Where a is first term or a1 and d is the common difference and n is any natural number 

When n = 10 : 

∴ a10 = a + (10 – 1)d 

⇒ a10 = a + 9d 

Similarly, 

When n = 18 : 

∴ a18 = a + (18 – 1)d 

⇒ a18 = a + 17d 

According to question : 

a10 = 41 and 

a18 = 73 

⇒ a + 9d = 41 ………………(i) 

And a + 17d = 73…………..(ii) 

Subtracting equation (i) from (ii) : 

a + 17d – (a + 9d) = 73 – 41 

⇒ a + 17d – a – 9d = 32 

⇒ 8d = 32

⇒ d = \(\frac{32}{8}\) 

⇒ d = 4 

Put the value of d in equation (i) : 

a + 9(4) = 41 

⇒ a + 36 = 41 

⇒ a = 41 – 36 

⇒ a = 5 

As, 

an = a + (n – 1)d 

a26 = a + (26 – 1)d 

⇒ a26 = a + 25d 

Now,

Put the value of a = 5 and 

d = 4 in a26 

⇒ a26 = 5 + 25(4) 

⇒ a26 = 5 + 100 

⇒ a26 = 105 

Hence, 

26th term of the given A.P. is 105.

578.

In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.

Answer»

For an A.P., let a be the first term and d be the common difference. 

t19 = 52, t38 = 128 …[Given] 

Since, tn = a + (n – 1)d 

∴ t19 = a + (19 – 1)d 

∴ 52 = a + 18d i.e. a + 18d = 52 …(i) 

Also, t38 = a + (38 – 1)d 

∴ 128 = a + 37d i.e. a + 37d = 128 …(ii) 

Adding equations (i) and (ii), we get

a + 18d = 52

(a + 37d = 128)/(2a + 55d = 180) .....(iii)

    Now, Sn = n/2 [ 2a + (n - 1)d]

∴ S56 = 56/2 [ 2a + (56 - 1)d]

= 28(2a + 55d)

= 28 x 180   ....[From (iii) ]

∴ S56 = 5040

∴ The sum of the first 56 terms is 5040.

579.

Insert six arithmetic means between 11 and -10.

Answer»

To find: Six arithmetic means between 11 and -10

Formula used: (i) d = b-a/n+1, where, d is the common difference

n is the number of arithmetic means

(ii) An = a + nd

We have 11 and -10

Using Formula, d = b-a/n+1

d = -10-(11)/6+1 = -21/7 = -3

Using Formula, An = a + nd

First arithmetic mean, A1 = a + d

= 11 + (-3) = 8

Second arithmetic mean, A2 = a + 2d

= 11 + 2(-3) = 11 + (-6) = 5

Third arithmetic mean, A3 = a + 3d

= 11 + 3(-3) = 11 + (-9) = 2

Fourth arithmetic mean, A4 = a + 4d

= 11 + 4(-3) = 11 + (-12) = -1

Fifth arithmetic mean, A5 = a + 5d

= 11 + 5(-3) = 11 + (-15) = -4

Sixth arithmetic mean, A6 = a + 6d

= 11 + 6(-3) = 11 + (-18) = -7

(Ans) The six arithmetic means between 11 and -10 are 8, 5, 2, - 1, -4 and -7.

580.

Insert three arithmetic means between 23 and 7.

Answer»

To find: Three arithmetic means between 23 and 7

Formula used: (i) d = b-a/n+1, where, d is the common difference

n is the number of arithmetic means

(ii) An = a + nd

We have 23 and 7

Using Formula, d = b - a/n+1

d = 7- 23/3+1 = -16/4 = - 4

Using Formula, An = a + nd

First arithmetic mean, A1 = a + d

= 23 + (- 4) = 19

Second arithmetic mean, A2 = a + 2d

= 23 + 2(- 4) = 23 + (-8) = 15

Third arithmetic mean, A3 = a + 3d

= 23 + 3(- 4) = 23 + (-12) = 11

(Ans) The three arithmetic means between 23 and 7 are 19, 15 and 11

581.

Insert six A.M.s between 15 and - 13.

Answer»

Let A1, A2, A3, A4, A5, A6 be the 7 AM Between 15 And - 13 

Then, 

15, A1, A2, A3, A4, A5, A6, - 13 are in AP 

We know, 

An = a + (n - 1)d 

A8 = - 13 = 15 + (8 - 1)d 

d = -4 

Hence, 

A1 = a + d 

= 15 - 4 

= 11 

A2 = A1 + d 

= 11 - 4 

= 7 

A3 = A2 + d 

= 7 - 4 

= 3 

A4 = A3 + d 

= 3 - 4 

= -1 

A5 = A4 + d 

= -1 - 4

= -5 

A6 = A5 + d 

= -5 - 4 

= -9

582.

Insert A.M.s between 7 and 71 in such a way that the 5 th A.M. is 27. Find the number of A.M.s.

Answer»

Let the series be 7, A1, A2, A3, ........, An, 71 

We know,

Total terms in AP are n + 2 

So,

71 is the (n + 2)th term 

We know, 

An = a + (n - 1)d 

So, 

A6 = a + (6 - 1)d 

a + 5d = 27 (5th term) 

d = 4 

71 = (n + 2)th term 

71 = a + (n + 2 - 1)d 

71 = 7 + n(4) 

n = 15 

There are 15 terms in AP.

583.

If demotes the sum of first n terms of an A.P., prove that S12 = 3 (s8 – S4) .

Answer»

S8 = \(\frac{8}{2}\)[2a + 7d] 

= 4 (2a + 7d).....  (i) 

S4 = \(\frac{4}{2}\)[2a + 7d] 

= 2 (2a + 7d)....... (ii) 

L.H.S = S12 = \(\frac{12}{2}\)[2a + 11d] 

= 6 [2a + 11d] 

R.H.S = 3 (S8 – S4

= 3 [4 (2a + 7d) – 2 (2a + 3d)] 

[From (i) and (ii)]

= 6 [4a + 14d – 2a – 3d] 

= 6 [2a + 11d] 

Since, L.H.S = R.H.S 

Hence, proved

584.

Find S2 for the following given A.P.  3, 5, 7, 9, ……. . 

Answer»

The given A.P. is 3, 5, 7, 9, ....

Here, t1 = 3, t2 = 5

Now, S2 = t1 + t2

\(\therefore\) S2 = 3 + 5 = 8

585.

For an A.P. t3 = 8 and t4 = 12, find the common difference d.

Answer»

Given, t3 = 8, t4 = 12

\(\therefore\) d = t4 – t3 = 12 – 8 = 4

586.

If S1 , S2 , S3 are the sums of first n natural numbers, their squares and their cubes respectively, then S3 (1 + 8S1 ) = A) 9S22B) S22C) 9S2D) 9S2

Answer»

Correct option is (A) \(9S_2\,^2\)

\(S_1=1+2+3+........+n=\frac{n(n+1)}2\)          ________________(1)

\(S_2=1^2+2^2+3^2+........+n^2=\frac{n(n+1)(2n+1)}6\)   ________________(2)

\(S_3=1^3+2^3+3^3+........+n^3=\left(\frac{n(n+1)}2\right)^2\)   ________________(3)

Now, \(S_3(1+8S_1)\) \(=\left(\frac{n(n+1)}2\right)^2\left(1+\frac{8n(n+1)}2\right)\)  (From (1) & (3))

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

\(=\frac{(n(n+1))^2}4\,\left(2n+1\right)^2\)

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

\(=\frac{(6S_2)^2}4\)                   (From (2))

\(=\frac{36S_2\,^2}4\) \(=9S_2\,^2\)

Correct option is A) 9S22

587.

Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.a, 2a, 3a, 4a, ....

Answer»

We have,

a2 – a1 = 2a – a = a

a3 – a2 = 3a – 2a = a

a4 – a3 = 4a – 3a = a

i.e. ak+1 – ak is the same every time.

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

Now, we have to find the next three terms.

We have a1 = a, a2 = 2a, a3 = 3a and a4 = 4a

Now, we will find a5, a6 and a7

So, a5 = 4a + a = 5a

a6 = 5a + a = 6a

and a7 = 6a + a = 7a

Hence, the next three terms are 5a, 6a and 7a

588.

Find the sum of first 15 terms of sequences having nth term as:xn = 6 – n

Answer»

Given an A.P. whose nth term is given by xn = 6 – n

To find the sum of the n terms of the given A.P., using the formula

Sn = \(\frac{n(a \,+\, l)}{2}\)

Where, a = the first term l = the last term.

Putting n = 1 in the given xn, we get

a = 6 – 1 = 5

For the last term (l), here n = 15

a15 = 6 – 15 = -9

So, Sn = \(\frac{15(5 \,–\, 9)}{2}\)

= 15 x (-2)

= -30

Therefore, the sum of the 15 terms of the given A.P. is S15 = -30

589.

Find the sum of first 20 terms the sequence whose nth term is an = An + B.

Answer»

Given an A.P. whose nth term is given by, an = An + B

We need to find the sum of first 20 terms.

To find the sum of the n terms of the given A.P., we use the formula,

Sn = \(\frac{n(a \,+\, l)}{2}\)

Where, a = the first term l = the last term,

Putting n = 1 in the given an, we get

a = A(1) + B = A + B

For the last term (l), here n = 20

A20 = A(20) + B = 20A + B

S20 = \(\frac{20}{2}\)((A + B) + 20A + B)

= 10[21A + 2B]

= 210A + 20B

Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B

590.

If S1 be the sum of (2n + 1) terms of an A.P. and S2 sum of its odd terms, then prove that: S1: S2 = (2n + 1) : (n + 1).

Answer»

To prove : S1: S2 = (2n + 1) : (n + 1)

We know that the sum of AP is given by the formula :

s = \(\frac{n}{2}\)(2a + (n-1)d)

Substituting the values in the above equation,

s1\(\frac{2n+1}{2}\)(2a+2nd)

For the sum of odd terms, it is given by,

s2 = a1 + a3 + a5 + …. a2n + 1 

s2 = a + a + 2d + a + 4d + … + a + 2nd 

s2 = (n + 1)a + n(n + 1)d 

s2 = (n + 1) (a + nd) 

Hence,

s: s\(\frac{2n+1}{n+1}\)

591.

If the first, second and last term of an A.P. are a, b and 2a respectively, its sum isA. \(\frac{ab}{2(b -a)}\)B . \(\frac{ab}{b-a}\)C. \(\frac{3ab}{2(b-a)}\)D. none of these

Answer»

a, b and 2a are in A.P so a is the first term and 2a is the last term denoted by T and Tn respectively. Here Common difference = b – a 

Tn = 2a = a + (n–1) (b–a) 

So n = \(\frac{b}{b-a}\)

Sum = \(\frac{n}{2}\){first term + last term}

\(\frac{b}{2(b-a)}\{3a\}\)

\(\frac{3ab}{2(b -a)}\)

592.

The sum of first n terms of an AP is (5n - n2). The nth term of the AP is(a) ( 5 - 2n) (b) ( 6 – 2n) (c) (2n – 5) (d) (2n – 6)

Answer»

Let Sn denotes the sum of first n terms of the AP.

∴ Sn = 5n - n2

\(\Rightarrow\) Sn-1 = 5(n - 1) - (n - 1)2

= 5n - 5 - n2 + 2n - 1

= 7n - n2 - 6

∴ nth term of the AP, an = Sn - sn-1

= (5n - n2) - (7n - n2 - 6)

Thus, the nth term of the AP is (6 - 2n).

593.

Find the sum of first 15 terms of sequences having nth term as:yn = 9 – 5n

Answer»

Given an A.P. whose nth term is given by yn = 9 – 5n

To find the sum of the n terms of the given A.P., using the formula,

Sn = \(\frac{n(a \,+\, l)}{2}\)

Where, a = the first term l = the last term.

Putting n = 1 in the given yn, we get

a = 9 – 5(1) = 9 – 5 = 4

For the last term (l), here n = 15

a15 = 9 – 5(15) = -66

So, Sn = \(\frac{15(4 \,–\, 66)}{2}\)

= 15 x (-31)

= -465

Therefore, the sum of the 15 terms of the given A.P. is S15 = -465

594.

If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to :A. \(\frac{1}{2}\)p3B. mn p C. p3D. (m+n)p2

Answer»

Option : (C)

Here, 

Sn = n2

Now, 

Substituting n = p we get 

Sp = p2

= p3

595.

If Sn = n2p and Sm = m2p,m≠n, in an A.P., prove that Sp = p3

Answer»

Given an AP whose sum of n terms is n2p and same AP with m terms whose sum is m2

To prove : Sp = p3 

We know that the sum of AP is given by the formula :

s = \(\frac{n}{2}\)(2a + (n-1)d)

Substituting the values in the above equation, we get

\(\frac{n}{2}\)(2a + (n-1)d) = n2…. (i)

Similarly, 

For series with m terms \(\frac{m}{2}\)(2a + (m-1)d) = m2p ...(ii)

Subtracting (ii) from (i) we get,

d = 2p 

Substituting d in (i) we get,

a = p 

Now,

Using the sum formula for AP consisting of p terms we get

Sp\(\frac{p}{2}\)(2a + (n-1)d)

Substituting the values in the above equation,

Sp\(\frac{p}{2}\)(2p + (p-1)2p)

SP = p3

596.

If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal toA. \(\frac{1}{2} p^3\)B. mn p C. p 3 D. (m + n) p2

Answer»

Correct answer is C. P3

Let first term = a and Common difference = d 

∴ According to the question, Sn = n2

Sn = n/2 (2a + (n–1) d) = n2

2a + (n–1) d = 2np……………….(1) 

And Sm = m2

Sm= m/2 (2a + (m–1) d) = m2

2mp = (2a + (m–1) d)……………….(2) 

Subtracting 2 from 1 

2a + (n–1) d – 2a – (m–1) d = 2np – 2mp 

d (n–1 –m + 1) = 2p (n– m) d = 2p 

putting value of d in (1) 

2a + (n–1) 2p = 2np 

a + (n–1)p = np 

a = p 

now Sp = p/2 (2a + (p–1) d) 

putting value of a = p and d = 2p 

Sp = p/2 (2p + (p–1) 2p) 

Sp = p3

597.

A sequence is defined by an = n3 – 6n2 + 11n – 6, n ∈ N. Show that the first three terms of the sequence are zero and all other terms are positive.

Answer»

Given,

an = n3 – 6n2 + 11n – 6, n ∈ N 

We can find first three terms of sequence by putting the values of n form 1 to 3. 

When n = 1 : 

a1 = (1)3 – 6(1)2 + 11(1) – 6 

⇒ a1 = 1 – 6 + 11 – 6 

⇒ a1 = 12 – 12 

⇒ a1 = 0 

When n = 2 : 

a2 = (2)3 – 6(2)2 + 11(2) – 6 

⇒ a2 = 8 – 6(4) + 22 – 6 

⇒ a2 = 8 – 24 + 22 – 6 

⇒ a2 = 30 – 30 

⇒ a2 = 0 

When n = 3 : 

a3 = (3)3 – 6(3)2 + 11(3) – 6 

⇒ a3 = 27 – 6(9) + 33 – 6 

⇒ a3 = 27 – 54 + 33 – 6 

⇒ a3 = 60 – 60 

⇒ a3 = 0 

This shows that the first three terms of the sequence is zero. 

When n = n : 

an = n3 – 6n2 + 11n – 6 

⇒ an = n3 – 6n2 + 11n – 6 – n + n – 2 + 2 

⇒ an = n3 – 6n2 + 12n – 8 – n + 2 

⇒ an = (n)3 – 3×2n(n – 2) – (2)3 – n + 2 

{(a – b)3 = (a)3 – (b)3 – 3ab(a – b)} 

⇒ an = (n – 2)3 – (n – 2) 

Here, 

n – 2 will always be positive for n > 3 

∴ an is always positive for n > 3.

598.

Deepak repays his total loan of Rs 1 ,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month. what amount will be paid as the last instalment of loan?(A) Rs 4900(B) Rs 5400(C) Rs 3500(D) Rs 4500

Answer»

The correct option is: (A) Rs 4900

Explanation:

1st instalment = Rs 1000

2nd instalment = Rs1000 + Rs 100 = Rs 1100

3rd instalment = Rs 1100 + Rs 100 = Rs 1200 and so on

Let number of instalments = n

.'. 1000 + 1100 + 1200 + ... up to n terms = 118000

=> n/2[2 x 1000 + (n - 1) 100] = 118000

=> 100n2 + 1900n - 236000 = 0

=> n2 + 19n - 2360 = 0

=> (n + 59)(n - 40) = 0

.'. Total no. of instalments = 40

Now, last instalment = 40th instalment

. .. a40 = a + 39d = Rs 4900

599.

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹ 250 for the second day, ₹300 for the third day, etc. the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Answer»

It is given that the penalty for each succeeding day is 50 more than for the preceding day, so the amount of penalties are in AP with common difference ₹50

Number of days in the delay of the work = 30

The amount of penalties are ₹200, ₹250, ₹300,... up to 30 terms.

∴ Total amount of money paid by the contractor as penalty,

S30 = ₹ 200 + ₹ 250 + ₹ 300 + ....... up to 30 terms

Here, a = ₹ 200, d = ₹ 50 and n = 30

Using the formula, \(S_n=\frac{n}{2}[2a+(n-1)d],\) we get

\(S_{30}=\frac{30}{2}[2\times200+(30-1)\times50]\)

= 15 (400 - 1450)

= 15 x 1850

= 27750

Hence, the contractor has to pay ₹27,750 as penalty

600.

If a = -1, d = -3 in an A.P, then 12th term is A) 17 B) -23 C) -34 D) -17

Answer»

Correct option is (C) -34

 \(a=-1,d=-3\)

\(\therefore a_{12}=a+11d\)

\(=-1+11\times-3\)

\(=-1-33=-34\)

\(\therefore\) \(12^{th}\) term is -34.

Correct option is C) -34