Question

Question 3 In the following APs find the missing term in the boxes
I. 2, __, 26
II. __, 13, __, 3
III. 5,__,__, 9.1/2
IV. -4, __, __, __, __, 6
V. __, 38, __,__,__,-22

Class 10 - Math - Arithmetic Progressions Page 106

Answer 1

(i) For this A.P.,

a = 2
a3 = 26
We know that, an = a + (n − 1) d
a3 = 2 + (3 - 1) d
26 = 2 + 2d
24 = 2d
d = 12
a2 = 2 + (2 - 1) 12
= 14
Therefore, 14 is the missing term.

(ii) For this A.P.,
a2 = 13 and
a4 = 3
We know that, an = a + (n − 1) d
a2 = a + (2 - 1) d
13 = a + d ... (i)
a4 = a + (4 - 1) d
3 = a + 3d ... (ii)
On subtracting (i) from (ii), we get
- 10 = 2d
d = - 5
From equation (i), we get
13 = a + (-5)
a = 18
a3 = 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,
a = 5 and
a4 = 19/2
We know that, an = a + (n − 1) d
a4 = a + (4 - 1) d
19/2 = 5 + 3d
19/2 - 5 = 3d3d = 9/2
d = 3/2

a2 = a + (2 - 1) d
a2 = 5 + 3/2
a2 = 13/2

a3 = a + (3 - 1) d

a3 = 5 + 2×3/2

a3 = 8

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,
a = −4 and
a6 = 6
We know that,
an = a + (n − 1) d
a6 = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 (2) = 0
a4 = a + 3d = − 4 + 3 (2) = 2
a5 = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v)
For this A.P.,
a2 = 38
a6 = −22
We know that
an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a + d ... (i)
a6 = a + (6 − 1) d
−22 = a + 5d ... (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d 
d = −15
a = a2 − d = 38 − (−15) = 53
a3 = a + 2d = 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

Answer 2

General form of an AP.:

a, a+d, a+2d, a+3d…….

Here a is the first term and d is common difference.

General term or nth term of A.P

The general term or nth term of A.P is given by an = a + (n – 1)d, 

where a is the first term, d is the common difference and n is the number of term

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Solution:

(i) Given:

a = 2
a3 = 26

 an = a + (n − 1) d

a3 = 2 + (3 – 1) d
26 = 2 + 2d
24 = 2d
d = 12

a2 = 2 + (2 – 1) 12
a2 = 14

Hence , 14 is the missing term.

(ii) Given:

a2 = 13 
a4 = 3

 an = a + (n − 1) d

a2 = a + (2 – 1) d
13 = a + d …......................................................... (i)

a4 = a + (4 – 1) d
3 = a + 3d …........................................................... (ii)

On subtracting (i) from (ii), 
– 10 = 2d
d = – 5

From equation (i), we get
13 = a + (-5)
a = 18

a3 = 18 + (3 – 1) (-5)
= 18 + 2 (-5) = 18 – 10 = 8

a3= 8

Hence, the missing terms are 18 and 8

(iii) Given:

a = 5
a4 = 9 1/2

 an = a + (n − 1) d

a4 = a + (4 – 1) d
19/2 = 5 +3 d 

3d= 19/2 - 5 = 9/2

3d = 9/2 

d= 9/2 ×1/3 = 3/2

d= 3/2

a2 = a+d

a2= 5+ 3/2 = 13/2

a2 = 13/2

a3 = a + 2 d
a3 = 5 +( 2 ×3/2 ) = 5 + 3=8

a3=8

Hence the missing terms are a2 = 13/2 &  a3 = 8

(iv) Given:

a = −4 and
a6 = 6

an = a + (n − 1) d

a6 = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2

a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 (2) = 0
a4 = a + 3d = − 4 + 3 (2) = 2
a5 = a + 4d = − 4 + 4 (2) = 4

Hence , the missing terms are −2, 0, 2, and 4 

(v)

Given:
a2 = 38
a6 = −22

an = a + (n − 1) d

a2 = a + (2 − 1) d
38 = a + d …................................................................(i)

a6 = a + (6 − 1) d
−22 = a + 5d ….......................................................... (ii)

On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15

on putting d = -15 in eq 1

38 = a + d 

38 = a+(-15)

38 = a-15

38+15 =a

a= 53

a3 = a + 2d = 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7

Hence  the missing terms are 53, 23, 8, and −7 

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hope this will help you.....