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501.

Find(i) 10th term of A.P. 2, 7, 12, ….(ii) 18th term of A.P. √2, 3√2, 5√2, …..(iii) 24th term of A.P. 9, 13, 17, 21, ….

Answer»

(i) Given A.P. 2, 7, 12, …..

First term a = 2

Common difference d = 7 – 2 = 5

nth term an = a + (n – 1)d

∴ a10 = 2 + (10 – 1) × 5

= 2 + 9 × 5

= 2 + 45

= 47

Hence, 10th term of given series 47.

(ii) Given A.P. √2, 3√2, 5√2, …..

First term a = √2

Common term d = 3√2 – √2

= √2(3 – 1) = 2√2

nth term an = a + (n – 1)d

∴ a18 = √2 + (18 – 1)2√2

= √2 + 17 × 2√2

= √2 + 34√2

= 35√2

Hence a18 = 35√2

(iii) Given A.P. 9, 13, 17, 21, …..

First term a = 9

common  difference

d = 13 – 9 = 4

nth term an = a + (n – 1)d

∴ a24 = 9 + (24 – 1) × 4

= 9 + 23 × 4

= 9 + 92

= 101

Hence a24 = 101

502.

The nth term of an A.P is given by (-4n + 15). Find the sum of first 20 terms of this A.P.

Answer»

Given,

The nth term of the A.P = (-4n + 15)

So, by putting n = 1 and n = 20 we can find the first ans 20th term of the A.P

a = (-4(1) + 15) = 11

And,

a20 = (-4(20) + 15) = -65

Now, for find the sum of 20 terms of this A.P we have the first and last term.

So, using the formula

Sn = \(\frac{n}{2}\)(a + l)

S20  = \(\frac{20}{2}\)(11 + (-65))

= 10(-54)

= -540

Therefore, the sum of first 20 terms of this A.P. is -540.

503.

Find the 15th term of the AP -40, -15, 10, 35, ……

Answer»

The 15th term of the AP -40, -15, 10, 35, ……

Given: AP is -40, -15, 10, 35, ……

Here, first term = a = -40

Common difference = d = -15 – (-40) = -15 + 40 = 25

n = 15

an = a + (n-1)d

a15 = -40 + (15 – 1)25

= 310

504.

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Answer»

Given,

24th term is twice the 10th term.

We know that, nth term an = a + (n – 1)d

⇒ a24 = 2(a10)

a + (24 – 1)d = 2(a + (10 – 1)d)

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a = 5d …. (1)

Now, the 72nd term can be expressed as

a72 = a + (72 – 1)d

= a + 71d

= a + 5d + 66d

= a + a + 66d [using (1)]

= 2(a + 33d)

= 2(a + (34 – 1)d)

= 2(a34)

⇒ a72 = 2(a34)

Hence, the 72nd term is twice the 34th term of the given A.P.

505.

Find:10th term of the AP -40, -15, 10, 35, …………. 

Answer»

Given A. P. is – 40, -15, 10, 35, ……….

First term (a) = -40

Common difference (d) = Second term – fast term

= -15 – (- 40)

= 40 – 15

= 25

We know that, nth term of an A.P. an = a + (n – 1)d

Then, 10th term of A. P. a10 = -40 + (10 – 1)25

= – 40 + 9.25

= – 40 + 225

= 185

∴ 10th term of the A. P. is 185.

506.

Find n if the given value of x is the nth term of the given A.P.1, \(\frac{21}{11}\), \(\frac{31}{11}\), \(\frac{41}{11}\), …; x = \(\frac{171}{11}\)

Answer»

Given A.P.  1, \(\frac{21}{11}\), \(\frac{31}{11}\), \(\frac{41}{11}\)\(\frac{171}{11}\)

Here, a = 1 d = \(\frac{21}{11}\) – 1 = \(\frac{10}{11}\)

Last term (nth term) = \(\frac{171}{11}\)

We know that an = a + (n – 1)d

\(\frac{171}{11}\) = 1 + (n-1)\(\frac{10}{11}\)

171 = 11 + 10n – 10

n = 170/10

n = 17

507.

If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.

Answer»

Let’s consider a to be the first term and d be the common difference.

And we know that, sum of first n terms is:

Sn = \(\frac{n}{2}\)(2a + (n − 1)d) and nth term is given by: an = a + (n – 1)d

Now, from the question we have

S10 = 120

⟹ 120 = \(\frac{10}{2}\)(2a + (10 − 1)d)

⟹ 120 = 5(2a + 9d)

⟹ 24 = 2a + 9d …. (1)

Also given that, a10 = 21

⟹ 21 = a + (10 – 1)d

⟹ 21 = a + 9d …. (2)

Subtracting (2) from (1), we get

24 – 21 = 2a + 9d – a – 9d

⟹a = 3

Now, on putting a = 3 in equation (2), we get

3 + 9d = 21

9d = 18

d = 2

Thus, we have the first term(a) = 3 and the common difference(d) = 2

Therefore, the nth term is given by

an  = a + (n – 1)d = 3 + (n – 1)2

= 3 + 2n -2

= 2n + 1

Hence, the nth term of the A.P is (an) = 2n + 1.

508.

The 17th term of an AP exceeds its 17th term by 7. Find the common difference.

Answer»

a17 = a10 + 7, d =? 

a + 16d = a + 9d + 7 

a + 16d – a – 9d = 7 

7d = 7

∴ d = \(\frac{7}{7}\)

∴ d = 1

509.

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.

Answer»

S7 = 63 

\(\frac{7}{2}\)[2(a) + 6d] = 639 

a + 3d = 9 (i)

a8 = a + 7d

Hence, for next 7 terms first term will be the 8th term i.e. a + 7d 

Sum of next 7 terms, S’7 = \(\frac{7}{2}\)[2(a + 7d) + 6d] 

161 = 7 [a + 7d + 3d] 

23 = a + 10d 

23 = 9 – 3d + 10d [From (i)] 

14 = 7d 

d = 2 

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

A = 9 – 3(2) = 3 

Now, a28= a + 27d 

= 3 + 27(2) 

= 3 + 54 

= 57

510.

The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.

Answer»

Let’s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

Sn = \(\frac{n}{2}\)(2a + (n − 1)d)

Given that sum of the first 7 terms of an A.P. is 63.

S7 = 63

And sum of next 7 terms is 161.

So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms

S14 = 63 + 161 = 224

Now, having

S7 = \(\frac{7}{2}\)(2a + (7 − 1)d)

⟹ 63(2) = 7(2a + 6d)

⟹ 9 × 2 = 2a + 6d

⟹ 2a + 6d = 18 . . . . (1)

And,

S14 = \(\frac{14}{2}\)(2a + (14 − 1)d)

⟹ 224 = 7(2a + 13d)

⟹ 32 = 2a + 13d …. (2)

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

⟹ 13d – 6d = 32 – 18

⟹ 7d = 14

⟹ d = 2

Using d in (1), we have

2a + 6(2) = 18

2a = 18 – 12

a = 3

Thus, from nth term

⟹ a28  = a + (28 – 1)d

= 3 + 27 (2)

= 3 + 54 = 57

Therefore, the 28th term is 57.

511.

Write the value of n for which nth of the A.P.s 3, 10, 17, ... and 63, 65, 67, ... are equal.

Answer»

Let nth term of A.P. 3,10,17,… be an 

And nth term of A.P. 63,65,67,…be an'

So,a1 = 3 and a2 = 10 

∴ d = a2 - a1 

= 10 - 3 

= 7 

And, 

a1' = 63 and a2' = 65 

∴ d' = a2' - a1' = 65 - 63 = 2 

Now, 

an = a1+(n-1)d 

= 3+(n-1)7 

= 3+7n-7 

= 7n-4 

And, 

an'=a1'+(n-1)d' 

= 63+(n-1)2 

= 63+2n-2 

= 2n+61 

Now,

It is given that an = an'

∴ 7n-4 = 2n+61 

∴ 7n-2n = 61+4 

∴ 5n = 65 

∴ n = 13

512.

For what value of n are the nth term of the following two A.P's the same. Also find this term 63, 65, 67, ... and 3, 10, 17, ....

Answer»

1st AP = 63, 65, 67, ...

Here, a = 63, d = 65 – 63 = 2

and 2nd AP = 3, 10, 17, ...

Here, a = 3, d = 10 – 3 = 7

According to the question,

63 + (n – 1)2 = 3 + (n – 1)7

⇒ 63 + 2n – 2 = 3 + 7n – 7

⇒ 61 + 2n = 7n – 4

⇒ 65 = 7n – 2n

⇒ 5n = 65

⇒ n = 13

13th term of the given AP’s are same.

Now, we will find the 13th term

We have,

an = a + (n – 1)d

a13 = 63 + (13 – 1)2

a13 = 63 + 12 × 2

a13 = 63 + 24

a13 = 87

513.

In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P.

Answer»

Let’s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

Sn = \(\frac{n}{2}\)(2a + (n − 1)d)

Given that sum of the first 10 terms of an A.P. is -150.

S10 = -150

And the sum of next 10 terms is -550.

So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms

S20 = -150 + -550 = -700

Now, having

S10 = \(\frac{10}{2}\)(2a + (10 − 1)d)

⟹ -150 = 5(2a + 9d)

⟹ -30 = 2a + 9d

⟹ 2a + 9d = -30 . . . . (1)

And,

S20 = \(\frac{20}{2}\)(2a + (20 − 1)d)

⟹ -700 = 10(2a + 19d)

⟹ -70 = 2a + 19d …. (2)

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

⟹ 19d – 9d = -70 – (-30)

⟹ 10d = -40

⟹ d = -4

Using d in (1), we have

2a + 9(-4) = -30

2a = -30 + 36

a = \(\frac{6}{2}\) = 3

Hence, we have a = 3 and d = -4

So, the A.P is 3, -1, -5, -9, -13,…..

514.

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

Answer»

Given,

First term of the A.P is 1505 and

S14 = 1505

We know that, the sum of first n terms is

Sn = \(\frac{n}{2}\)(2a + (n − 1)d)

So,

S14 = \(\frac{14}{2}\)(2(10) + (14 − 1)d) = 1505

7(20 + 13d) = 1505

20 + 13d = 215

13d = 215 – 20

d = \(\frac{195}{13}\)

d =15

Thus, the 25th term is given by

a25 = 10 + (25 -1)15

= 10 + (24)15

= 10 + 360

= 370

Therefore, the 25th term of the A.P is 370.

515.

For what value of n’. are the nth terms of two AP: 63, 65, 67, ……. and 3, 10, 17, ……….. equal?

Answer»

63, 65, 67,………. 

a = 63. d = 65 – 63 = 2. an =? 

nth term of this is 

an = a + (n – 1)d 

= 63 + (n – 1) 2 

= 63 + 2n – 2 

an= 2n + 61 …………….. (i) 

3, 10, 17, …………. 

a = 3, d = 10 – 3 = 7, an =? 

an = a + (n – 1)d 

= 3 + (n – 1)7 

= 3 + 7n – 7 

an = 7n — 4 ………….(ii) 

Here. nth terms of second AP are equal. 

∴ equation (i) = equation (ii) 

2n + 61 = 7n – 4 

2n – 7n = -4 – 61 5n = 65 

5n =65

∴ n = \(\frac{65}{5}\)

∴ n = 13 

∴13th terms of the two given AP are equal.

516.

Write the Properties of A.P.

Answer»

(A) For any real numbers a and b, the sequence whose nth term is an = an + b is always an A.P. with common difference ‘a’ (i.e. coefficient of term containing n)

(B) If any nth term of sequence is a linear expression in n then the given sequence is an A.P.

(C) If a constant term is added to or subtracted from each term of an A.P. then the resulting sequence is also an A.P. with the same common difference.

(D) If each term of a given A.P. is multiplied or divided by a non-zero constant K, then the resulting sequence is also an A.P. with common difference Kd or......... respectively. Where d is the common difference of the given A.P.

(E) In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of 1st and last term.

(F) If three numbers a,b,c are in A.P., then 2b = a + c.

517.

Check whether an = 2n2 + 1 is an A.p. or not.

Answer»

an = 2n2 + 1 Then an+1 = 2 (n + 1)2 + 1

So, an+1 - an = 2(n2 + 2n + 1) + 1 - 2n2 - 1

= 2n2 + 4n + 2 + 1 - 2n2 - 1

= 4n + 2, which is not constant 

So, The above sequence is not an A.P.

518.

Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years.

Answer»

Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm 

Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P. 

Here, a = 70, d = 80 – 70 = 10 

We have to find height of Kabir at the age of 15years i.e. t15

Now, tn = a + (n – 1)d 

∴ t15 = 70 + (15 – 1)10 

= 70 + 14 × 10 = 70 + 140 

∴ t15 = 210 

∴ The height of Kabir at the age of 15 years will be 210 cm.

519.

The 11th term of the AP: –5, (–5/2), 0, 5/2, …is(A) –20 (B) 20 (C) –30 (D) 30

Answer»

(B) 20

Explanation:

First term, a = – 5

Common difference,

d = 5 – (-5/2) = 5/2

n = 11

We know that the nth term of an AP is

an = a + (n – 1)d

Where,

a = first term

an is nth term

d is the common difference

a11 = – 5 + (11 – 1)(5/2)

a11 = – 5 + 25 = 20

520.

The first four terms of an AP, whose first term is –2 and the common difference is –2, are(A) – 2, 0, 2, 4(B) – 2, 4, – 8, 16(C) – 2, – 4, – 6, – 8(D) – 2, – 4, – 8, –16

Answer»

(C) – 2, – 4, – 6, – 8

Explanation:

First term, a = – 2

Second Term, d = – 2

a1 = a = – 2

We know that the nth term of an AP is

an = a + (n – 1)d

Where,

a = first term

an is nth term

d is the common difference

Hence, we have,

a2 = a + d = – 2 + (- 2) = – 4

Similarly,

a3 = – 6

a4 = – 8

So the A.P is

– 2, – 4, – 6, – 8

521.

Find the sum of the first (i) 11 terms of the A.P : 2,6,10,14, ....(ii) 13 terms of the A.P : -6,0,6,12,....(iii) 51 terms of the A.P. whose second term is 2 and fourth term is 8.

Answer»

(i)

a = 2, d = 6 – 2 = 4 

S11=\(\frac{11}{2}\) [2(a) + 10d]

\(\frac{11}{2}\)[2(2) + 10(4)]

= 11 [2 + 20]

= 242

(ii)

a = -6, d = 0 + 6 = 6 

S13= \(\frac{13}{2}\)[2(a) + 12d] 

= 13 [-6 + 6(6)]

= 13 [-6 + 36] 

= 13 (30) = 390

(iii)

a2= 2 a + d = 2 (i) 

a4 = 8 a + 3d = 8 

2 – d + 3d = 8 

2 + 2d = 8 

d = 3 

a = -1

\(S_{21} = \frac{51}{2}[2(a) + 50d]\)

= 51 [-1 + 25(3)] 

= 51 (74) 

= 3774

522.

Find the sum of the first:11 terms of the A.P. : 2, 6, 10, 14,  . . . 

Answer»

We know that the sum of terms for different arithmetic progressions is given by

Sn = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Given A.P 2, 6, 10, 14,… to 11 terms.

Common difference (d) = a2 – a1 = 10 – 6 = 4

Number of terms (n) = 11

First term for the given A.P. (a) = 2

So,

S11  = \(\frac{11}{2}\)[2(2) + (11 − 1)4]

\(\frac{11}{2}\)[2(2) + (10)4]

\(\frac{11}{2}\)[4 + 40]

= 11 × 22

= 242

Hence, the sum of first 11 terms for the given A.P. is 242

523.

Write first four terms of the AP, when the first term ‘a’ and the common difference’d’ are given as follows: (i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = -3 (iv) a = -1, d = (v) a = -1.25, d = 0.25

Answer»

(i) a = 10, d = 10 

First four terms are, 

a, a + d, a + 2d, a + 3d 

10, 10 + 10, 10 + 20, 10 + 30 

10. 20, 30, 40. 

(ii) a = -2, d = 0 

First four terms are, 

a, a + d, a + 2d, a + 3d 

-2. -2 + 0, -2 + 0, -2 + 0 

-2. -2. -2, -2

(iii) a = 4, d = -3 

First four terms are, 

a, a + d, a + 2d, a + 3d 

4, 4 – 3, 4 – 2 × 3, 4 – 3 ×3 

4, 1, -2, -5 

iv) a = -1, d = \(\frac{1}{2}\)

First four terms are, 

a, a + d, a + 2d, a + 3d 

- 1, - 1 + \(\frac{1}{2}, \) - 1 + 2 x \(\frac{1}{2}\), - 1 + 3 x \(\frac{1}{2}\)

- 1, - \(\frac{1}{2}, \) 0, \(\frac{1}{2}, \)

(v) a = -1.25, d = 0.25 

First four terms are, 

a, a + d, a + 2d, a + 3d 

- 1.25, - 1.25 + 0.25, - 1.25 + 2 × 0.25, - 1.25 + 3 × 0.25 

- 1.25, -1, - 0.75, - 0.50

524.

How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero? 

Answer»

Given A.P. is 27, 24, 21. . .

We know that,

Sn = \(\frac{n}{2}\)[2a + (n − 1)d]

Here we have, the first term (a) = 27

The sum of n terms (Sn) = 0

Common difference of the A.P. (d) = a2 – a1 = 24 – 27 = -3

On substituting the values in Sn, we get

⟹ 0 = \(\frac{n}{2}\)[2(27) + (n − 1)( − 3)]

⟹ 0 = (n)[54 + (n – 1)(-3)]

⟹ 0 = (n)[54 – 3n + 3]

⟹ 0 = n [57 – 3n] Further we have, n = 0 Or, 57 – 3n = 0

⟹ 3n = 57

⟹ n = 19

The number of terms cannot be zero,

Hence, the numbers of terms (n) is 19.

525.

How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693? 

Answer»

Given A.P. is 63, 60, 57,…

We know that,

Sn = \(\frac{n}{2}\)[2a + (n − 1)d]

Here we have,

the first term (a) = 63

The sum of n terms (Sn) = 693

Common difference of the A.P. (d) = a2 – a1 = 60 – 63 = –3

On substituting the values in Swe get

⟹ 693 = \(\frac{n}{2}\)[2(63) + (n − 1)(−3)]

⟹ 693 = \(\frac{n}{2}\)[163+(−3n + 3)]

⟹ 693 = \(\frac{n}{2}\)[129 − 3n]

⟹ 693(2) = 129n – 3n2

Now, we get the following quadratic equation.

⟹ 3n2 – 129n + 1386 = 0

⟹ n2 – 43n + 462

Solving by factorisation method, we have

⟹ n2 – 22n – 21n + 462 = 0

⟹ n(n – 22) -21(n – 22) = 0

⟹ (n – 22) (n – 21) = 0

Either, n – 22 = 0 ⟹ n = 22

Or,  n – 21 = 0 ⟹ n = 21

Now, the 22nd term will be a22 = a1 + 21d = 63 + 21( -3 ) = 63 – 63 = 0

So, the sum of 22 as well as 21 terms is 693.

Therefore, the number of terms (n) is 21 or 22.

526.

(5 + 13 + 21 + … + 181) = ? (a) 2476 (b) 2337 (c) 2219(d) 2139

Answer»

Correct answer is (d) 2139

Here, a = 5, d = (13 - 5) = 8 and l = 181

Let the number of terms be n.

Then, Tn = 181

\(\Rightarrow\) a + (n - 1) d = 181

\(\Rightarrow\) 5 + (n - 1)x8 = 181

\(\Rightarrow\) 8n = 184

\(\Rightarrow\) n = 23

∴ Required sum = \(\frac{n}{2}(a+l)\)

\(\frac{23}{2}(5+181)=23\times93=2139\)

Hence, the required sum is 2139.

527.

Find the sum of first n natural numbers.

Answer»

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 2 - 1 = 1 

To find : the sum of given AP 

The formula for the sum of AP is given by,

s = \(\frac{n}{2}\)(2a+(n-1)d)

Substituting the values in the above formula,

s = \(\frac{n}{2}\)(2+(n-1)(1))

s = \(\frac{n(n+1)}{2}\)

528.

Find the sum of first n odd natural numbers.

Answer»

Given,

An AP of first n odd natural numbers whose first term a is 1, and common difference d is 3 

Given,

Sequence is 1, 3, 5, 7……n 

To find : the sum of first n natural numbers 

Hence,

The sum is given by the formula s = \(\frac{n}{2}\)(2a+(n-1)d)

Substituting the values in the above equation we get,

 s = \(\frac{n}{2}\)(2+(n-1)2)

s = n2

529.

If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Answer»

Solution: Given, a3 = 4 and a9 = - 8

a3 = a + 2d = 4

a9 = a + 8d = - 8

Subtracting 3rd term from 9th term, we get;

a + 8d – a – 2d = - 8 – 4 = - 12

Or, 6d = - 12

Or, d = - 2

Substituting the value of d in 3rd term, we get;

a + 2(-2) = 4

Or, a – 4 = 4

Or, a = 8

Now; 0 = a + (n – 1)d

Or, 0 = 8 + (n – 1)(- 2)

Or, (n – 1)(- 2) = - 8

Or, n – 1 = 4

Or, n = 5

Thus, 5th term of this AP is zero.

530.

Find the sum of all natural numbers between a and 100, which are divisible by 3.

Answer»

a = 3, 

l = 99, 

n = 33

S33 = \(\frac{33}{2} (a + l)\)

\(\frac{33}{2} (3 + 99)\)

\(\frac{33}{2}\) (3 + 99)

\(\frac{33}{2}\) (102)

= 33 (51) = 1683

531.

If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Answer»

Given sum up to n th term Sn= 4n-n^2

So   n th term wil be 

tn = Sn - Sn-1= 4n -n^2-(4(n-1)-(n-1)^2

=> tn= 4+2n -1=5-2n

So sum of first 2 term

S2= 4×2-2^2=4

t1 = 5-2×1=3

t2= 5- 2×2=1

t3 = 5-2×3=-1

t10 = 5 - 2×10=-15

532.

Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149.

Answer»

a22= a + 21d 

149 = a + 21(22)

a = 149 – 462 

a = -313

S22\(\frac{22}{2}\) [2a + (22 - 1)d]

= 11 [-626 + 462]

= 11 (-164)

= -1804

533.

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Answer»

d = 7, a22 = 149, S22 = ? 

a = a + (n – 1)d 

a + (22 – 1)7 = 149 

a + 21 × 7 = 149 

a + 147 = 149 

=149 – 147

∴ a = 2

∴ Sn\(\frac{n}{2}\)[a + an]

S22 = \(\frac{22}{2}\)[2 + 149]

= 11[151] 

∴ S22 = 1661

534.

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Answer»

Solution:

Given d = 7,
22nd term = 149 , 
⇒ a+21d = 149
⇒a+21(7) = 149
⇒a+147 = 149
⇒a=149-147=2
Sn = (n/2)(2a+(n-1)d)
S₂₂ = (22/2)(2×2+(22-1)7)
S₂₂ = 11(4+21(7))
S₂₂ = 11(4+147)
S₂₂ = 11(151)
S₂₂ = 1661.

535.

Define an arithmetic progression.

Answer»

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP.

536.

Find the 20th term from the Last term of the AP: 3, 8, 13, …………., 253.

Answer»

3, 8, 13, ………….., 253 

a = 3. d = 8 – 3 = 5, an = 253 

20th term from the last term of the AP starting from 253 =? 

253, 258, 263, ………… a20 =? 

a = 253, d = 258 – 253 = 5, n = 20 

an = a + (n – 1)d 

a20= 253 + (20 – 1) 5 

= 253 + 19 × 5 

= 253 + 95 

∴ a20 = 348 

∴ 20th term from the last term of the AP is 348.

537.

Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149.

Answer»

Let the first term be taken as a.

Given,

a22 = 149 and the common difference d = 22

Also, we know that

an = a + ( n – 1) d

So, the 22nd term is given by

a22 = a + (22 – 1)d

149 = a + (21) (22)

a = 149 – 462

a = – 313

Now, for the sum of term

Sn = \(\frac{n}{2}\)[2a + (n − 1)d]

Here, n = 22

S22  = \(\frac{22}{2}\)[2(−313) + (22 − 1)(22)]

= (11)[ – 626 + 462]

= (11)[–164]

= – 1804

Hence, the sum of first 22 terms for the given A.P. is S22 = -1804

538.

The nth term of an A.P., the sum of whose n terms is Sn, is A. Sn + Sn–1 B. Sn – Sn–1 C. Sn + Sn + 1 D. Sn – Sn + 1

Answer»

The sum of n terms of an A.P is given by Sn 

Sn = \(\frac{n}{2}\)(2a + (n–1) d)

 If the sum of n terms is given that is Sn is given then the nth term is given by the formula 

Tn = Sn– Sn–1 Where Sn–1 is sum of the (n–1)th term of the A.P.

539.

Write the common difference of an A.P. whose nth term is an = 3n + 7.

Answer»

If an = 3n + 7 Then 

a1 = 3 (1) + 7 = 10 

a2 = 3(2) + 7 = 13 

a3 = 3(3) + 7 = 16

d= common difference = an – an–1 

= a3 – a2 

= 16 – 13 = 3

540.

30th term of the AP: 10, 7, 4, ……….. is (A) 97 (B) 77 (C) -77 (D) -87

Answer»

Answer is (C)

a = 10. d = 7 – 10 = -3, n = 30, a30 =? 

an = a + (n – 1)d 

a30 = 10 + (30 – 1)(-3) 

= 10 + 29(-3) 

= 10 – 87 

∴ a30 = -77

541.

Write the nth term of an A.P. the sum of whose n terms is Sn.

Answer»

First term = a 

Sum up to first term = a 

Last term (nth term) = an

Sum up to n terms = Sn 

Second last term = an–1 

Sum up to (n–1)th term = sn–1 

Therefore, an = Sn–Sn–1

542.

Which item of the sequence 114, 109, 104, .... is the first negative term?

Answer»

Here a = 114, d is common difference 

d = a3 – a2 = a2 – a1 = –5 

For finding first negative term Tn < 0 

a + (n – 1) d < 0 

114 + (n – 1) (– 5) < 0 

114 – 5n + 5 < 0 

119 – 5n < 0 – 5n < – 119 

5n > 119 

n > 119/5 

n > 23.8 

Therefore first negative term is 24th term

543.

If the nth term of the A.P. 9, 7, 5, …. is same as the nth term of the A.P. 15, 12, 9, … find n.

Answer»

Given,

A.P= 9, 7, 5, …. and A.P= 15, 12, 9, …

And, we know that, nth term an = a + (n – 1)d

For A.P1,

a = 9, d = Second term – first term = 9 – 7 = -2

And, its nth term a= 9 + (n – 1)(-2) = 9 – 2n + 2

a= 11 – 2n …..(i)

Similarly, for A.P2

a = 15, d = Second term – first term = 12 – 15 = -3

And, its nth term a= 15 + (n – 1)(-3) = 15 – 3n + 3

a= 18 – 3n …..(ii)

According to the question, its given that

nth term of the A.P1 = nth term of the A.P2

⇒ 11 – 2n = 18 – 3n

n = 7

Therefore, the 7th term of the both the A.Ps are equal.

544.

Find the 12th term from the end of the arithmetic progressions:1, 4, 7, 10, … ,88

Answer»

Given A.P = 1, 4, 7, 10, … ,88

Here, a = 1 and d = (4 – 1) = 3

Now, find the number of terms when the last term is known i.e, 88

an = 1 + (n – 1)3 = 88

1 + 3n – 3 = 88

3n = 90

n = 30

Hence, the A.P has 30 terms.

So, the 12th term from the end is same as (30 – 12 + 1)th of the A.P which is the 19th term.

⇒ a89 = 1 + (19 – 1)3

= 1 + 18(3)

= 1 + 54

= 55

Therefore, the 12th term from the end of the A.P is 55.

545.

Find the 12th term from the end of the arithmetic progressions:3,8,13, … ,253

Answer»

Given A.P = 3,8,13, … ,253

Here, a = 3 and d = (8 – 3) = 5

Now, find the number of terms when the last term is known i.e, 253

an = 3 + (n – 1)5 = 253

3 + 5n – 5 = 253

5n = 253 + 2 = 255

n = 255/5

n = 51

Hence, the A.P has 51 terms.

So, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term.

⇒ a40 = 3 + (40 – 1)5

= 3 + 39(5)

= 3 + 195

= 198

Therefore, the 12th term from the end of the A.P is 198.

546.

Write the value of a30 – a10 for the A.P. 4, 9, 14, 19, ……….

Answer»

Here a1 = 4 

a2 = 9 

a3 = 14 

d = common difference = a3 – a2 = a2 – a1 = 14 – 9 = 9 – 4 = 5 

a30 = a + (n–1) d 

a30 = 4 + (30 –1) 5 

a30= 4 + 29 X 5 

a30= 149

Similarly, a10 = 4 + 9x5

 a10= 49 

a30 – a10 = 149 – 49 = 100

547.

Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.

Answer»

Terms are in A.P. if common difference (d) is same between two continuous numbers. 

x + 10 – 2x = 3x + 2 – (x + 10) 

10 – x = 2x – 8 

18 = 3x 

x= 6 

So the terms in A.P. are: 2(6), 6 + 10, 3(6) + 2 

= 12, 16, 20

548.

Find the 12th term from the end of the arithmetic progressions:3, 5, 7, 9, …. 201

Answer»

Given A.P = 3, 5, 7, 9, …. 201

Here, a = 3 and d = (5 – 3) = 2

Now, find the number of terms when the last term is known i.e, 201

an = 3 + (n – 1)2 = 201

3 + 2n – 2 = 201

2n = 200

n = 100

Hence, the A.P has 100 terms.

So, the 12th term from the end is same as (100 – 12 + 1)th of the A.P which is the 89th term.

⇒ a89 = 3 + (89 – 1)2

= 3 + 88(2)

= 3 + 176

= 179

Therefore, the 12th term from the end of the A.P is 179.

549.

The 8th term of an A.P. is 37, and its 12th term is 57. Find the A.P.

Answer»

We have

a8 = a + (8 – 1)d = a + 7d = 37 …(1)

and a12 = a + (12 – 1)d = a + 11d = 57 …(2)

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

a + 7d – a – 11d = 37 – 57

⇒ – 4d = –20

⇒ d = 5

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

a + 7(5) = 37

⇒ a + 35 = 37

⇒ a = 2

Hence, the required AP is 2, 7, 12, 17,…

550.

Write 5th term from the end of the A.P. 3, 5, 7, 9, ...., 201.

Answer»

Here a = 201, 

a2 = 5, 

a3 = 7 

d = a3 – a2 = a2 – a1 

= 7 – 5 = 5 – 3 = 2 

tn = a + (n –1)d 

tn = 3 + (n–1)2 = 201 

(n –1)2 = 201 – 3 = 198 

n – 1 =\(\frac{198}{2}\) = 99 

n = 99 + 1 =100 

5th term from end = 96th term 

T95 = 3 + (96 –1)2 

T95 = 3 + 95 x 2 

T95 = 3 + 190 = 193