CBSE 10th Maths - arithmetic - progressions - Exercise 5.2

Solution:

(i) Given,

First term, a = 7

Common difference, d = 3

Number of terms, n = 8,

We have to find the $N^{th}$ term, $a_n$ = ?

As we know, for an A.P.,

$a_n$ = a+(n−1)d

Putting the values,

7+(8 −1) 3

7+(7) 3

7+21 = 28

Hence, $a_n$ = 28

(ii) Given,

First term, a = -18

Common difference, d = ?

Number of terms, n = 10

$N^{th}$ term, $a_n$ = 0

As we know, for an A.P.,

$a_n$ = a+(n−1)d

Putting the values,

0 = − 18 +(10−1)d

18 = 9d

d = $\dfrac{18}{9}$ = 2

Hence, common difference, d = 2

(iii) Given,

First term, a = ?

Common difference, d = -3

Number of terms, n = 18

$N^{th}$ term, $a_n$ = -5

As we know, for an A.P.,

$a_n$ = a+(n−1)d

Putting the values,

−5 = a+(18−1) (−3)

−5 = a+(17) (−3)

−5 = a−51

a = 51−5 = 46

Hence, a = 46

(iv) Given,

First term, a = -18.9

Common difference, d = 2.5

Number of terms, n = ?

$N^{th}$ term, $a_n$ = 3.6

As we know, for an A.P.

$a_n$ = a +(n −1)d

Putting the values,

3.6 = − 18.9+(n −1)2.5

3.6 + 18.9 = (n−1)2.5

22.5 = (n−1)2.5

(n – 1) = $\dfrac{22.5}{2.5}$

n – 1 = 9

n = 10

Hence, n = 10

(v) Given,

First term, a = 3.5

Common difference, d = 0

Number of terms, n = 105

$N^{th}$ term, $a_n$ = ?

As we know, for an A.P.

$a_n$ = a+(n −1)d

Putting the values,

$a_n$ = 3.5+(105−1) 0

$a_n$ = 3.5+104×0

$a_n$ = 3.5

Hence, $a_n$ = 3.5

Solution:

(i) Answer - Option C

Given here,

A.P. = 10, 7, 4, …

Therefore, we can find,

First term, a = 10

Common difference, $d = a_2 − a_1 = 7−10 = −3$

As we know, for an A.P.,

$a_n$ = a +(n−1)d

Putting the values;

$a_{30}$ = 10+(30−1)(−3)

$a_{30}$ = 10+(29)(−3)

$a_{30}$ = 10−87 = −77

Hence, the correct Answer is option C.'

(ii) Answer : Option B

Given here,

A.P. = -3, $-\dfrac{1}{2}$, ,2 …

Therefore, we can find,

First term a = – 3

Common difference, $d = a_2 − a_1 = (-\dfrac{1}{2}) -(-3)$

$(-\dfrac{1}{2}) + 3 = \dfrac{5}{2}$

As we know, for an A.P.,

$a_n$ = a+(n−1)d

Putting the values;

$a_{11} = -3+(11-1)(\dfrac{5}{2})$

$a_{11} = -3+(10)(\dfrac{5}{2})$

$a_{11} = -3+25$

$a_{11} = 22$

Hence, the Answer is option B.

Solution:

(i) For the given A.P : 2,__, 26

The first and third term are;

a = 2

$a_3$ = 26

As we know, for an A.P.

$a_n = a+(n −1)d$

Therefore, putting the values here,

$a_3$ = 2+(3-1)d

26 = 2+2d

24 = 2d

d = 12

$a_2$ = 2+(2-1)12 = 14

Therefore, 14 is the missing term.

(ii) For the given A.P : __ , 13,__ ,3

$a_2$ = 13 and

$a_4$ = 3

As we know, for an A.P.,

$a_n = a+(n−1) d$

Therefore, putting the values here,

$a_2$ = a +(2-1)d

13 = a+d ………………. (i)

$a_4$ = a+(4-1)d

3 = a+3d ………….. (ii)

On subtracting equation (i) from (ii), we get,

– 10 = 2d

d = – 5

From equation (i), putting the value of d,we get

13 = a+(-5)

a = 18

$a_3$ = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8

Therefore, the missing terms are 18 and 8 respectively.

(iii) For the given A.P : 5, __, __, 9$\dfrac{1}{2}$

a = 5 and

$a_4 = \dfrac{19}{2}$

As we know, for an A.P.,

$a_n = a+(n −1)d$

Therefore, putting the values here,

$a_4$ = a+(4-1)d

$\dfrac{19}{2}$ = 5+3d

$(\dfrac{19}{2})$ – 5 = 3d

3d = $\dfrac{9}{2}$

d = $\dfrac{3}{2}$

$a_2$ = a+(2-1)d

$a_2 = 5+\dfrac{3}{2}$

$a_2 = \dfrac{13}{2}$

$a_3 = a+(3-1)d$

$a_3 = 5+2×\dfrac{3}{2}$

$a_3$ = 8

Therefore, the missing terms are $\dfrac{13}{2}$ and 8 respectively.

(iv) For the given A.P : -4, __, __, __, __, 6

a = −4 and

$a_6$ = 6

As we know, for an A.P.,

$a_n = a+(n −1)d$

Therefore, putting the values here,

$a_6$ = a+(6−1)d

6 = − 4+5d

10 = 5d

d = 2

$a_2$ = a+d = − 4+2 = −2

$a_3$ = a+2d = − 4+2(2) = 0

$a_4$ = a+3d = − 4+ 3(2) = 2

$a_5$ = a+4d = − 4+4(2) = 4

Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v) For the given A.P : __,38,__,__,__22

$a_2$ = 38

$a_6$ = −22

As we know, for an A.P.,

$a_n = a+(n −1)d$

Therefore, putting the values here,

$a_2$ = a+(2−1)d

38 = a+d ……………………. (i)

$a_6$ = a+(6−1)d

−22 = a+5d …………………. (ii)

On subtracting equation (i) from (ii), we get

− 22 − 38 = 4d

−60 = 4d

d = −15

a = $a_2$ − d = 38 − (−15) = 53

$a_3$ = a + 2d = 53 + 2 (−15) = 23

$a_4$ = a + 3d = 53 + 3 (−15) = 8

$a_5$ = a + 4d = 53 + 4 (−15) = −7

Therefore, the missing terms are 53, 23, 8, and −7 respectively.

Solution:

Given the A.P. series as 3, 8, 13, 18, …

First term, a = 3

Common difference, $d = a_2 − a_1 = 8 − 3 = 5$

Let the nth term of given A.P. be 78. Now as we know,

$a_n = a+(n−1)d$

Therefore,

78 = 3+(n −1)5

75 = (n−1)5

(n−1) = 15

n = 16

Hence, $16^{th}$ term of this A.P. is 78.

Solution:

(i) Given, 7, 13, 19, …, 205 is the A.P

Therefore

First term, a = 7

Common difference, $d = a_2 − a_1$ = 13 − 7 = 6

Let there are n terms in this A.P.

$a_n = 205$

As we know, for an A.P.

$a_n = a + (n − 1) d$

Therefore, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

(ii) 18,15 $\dfrac{1}{2}$, 13, . . . , – 47

First term, a = 18

Common difference, $d = a_2-a_1 $

d = 5 $\dfrac{1}{2}$ - 18

d = $\dfrac{(31-36)}{2} = -\dfrac{5}{2}$

Let there are n terms in this A.P.

$a_n = -47$

As we know, for an A.P.

$a_n = a+(n−1)d$

-47 = 18+(n-1)(-$\dfrac{5}{2})$

-47-18 = (n-1)(-$\dfrac{5}{2})$

-65 = (n-1)(-$\dfrac{5}{2})$

(n-1) = -$\dfrac{130}{-5}$

(n-1) = 26

n = 27

Therefore, this given A.P. has 27 terms in it.

Solution:

For the given series, A.P. 11, 8, 5, 2..

First term, a = 11

Common difference, $d = a_2−a_1$ = 8−11 = −3

Let −150 be the $n^{th}$ term of this A.P.

As we know, for an A.P.,

$a_n = a+(n−1)d$

-150 = 11+(n -1)(-3)

-150 = 11-3n +3

-164 = -3n

n = $\dfrac{164}{3}$

Clearly, n is not an integer but a fraction.

Therefore, – 150 is not a term of this A.P.

Solution:

Given that,

$11^{th}$ term, $a_{11}$ = 38

and $16^{th}$ term, $a_{16}$ = 73

We know that,

$a_n$ = a+(n−1)d

$a_{11}$ = a+(11−1)d

38 = a+10d ………………………………. (i)

In the same way,

$a_{16}$ = a +(16−1)d

73 = a+15d ………………………………………… (ii)

On subtracting equation (i) from (ii), we get

35 = 5d

d = 7

From equation (i), we can write,

38 = a+10×(7)

38 − 70 = a

a = −32

$a_{31}$ = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, $31^{st}$ term is 178

Solution:

Given that,

$3^{rd}$ term, $a_3$ = 12

$50^{th}$ term, $a_{50}$ = 106

We know that,

$a_n = a+(n−1)d$

$a_3$ = a+(3−1)d

12 = a+2d ……………………………. (i)

In the same way,

$a_{50}$ = a+(50−1)d

106 = a+49d …………………………. (ii)

On subtracting equation (i) from (ii), we get

94 = 47d

d = 2 = common difference

From equation (i), we can write now,

12 = a+2(2)

a = 12−4 = 8

$a_{29}$ = a+(29−1) d

$a_{29}$ = 8+(28)2

$a_{29}$ = 8+56 = 64

Therefore, $29^{th}$ term is 64.

Solution:

Given that,

$3^{rd}$ term, $a_3$ = 4

and $9^{th}$ term, $a_9$ = −8

We know that,

$a_n = a+(n−1)d$

Therefore,

$a_3$ = a+(3−1)d

4 = a+2d ……………………………………… (i)

$a_9$ = a+(9−1)d

−8 = a+8d ………………………………………………… (ii)

On subtracting equation (i) from (ii), we will get here,

−12 = 6d

d = −2

From equation (i), we can write,

4 = a+2(−2)

4 = a−4

a = 8

Let $n^{th}$ term of this A.P. be zero.

$a_n = a+(n−1)d$

0 = 8+(n−1)(−2)

0 = 8−2n+2

2n = 10

n = 5

Hence, $5^{th}$ term of this A.P. is 0.

Solution:

We know that, for an A.P series;

$a_n = a+(n−1)d$

$a_{17}$ = a+(17−1)d

$a_{17}$ = a +16d

In the same way,

$a_{10}$ = a+9d

As it is given in the question,

$a_{17} − a_{10}$ = 7

Therefore,

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.