Arithmetic Progression

NCERT Exercise 5.2

Part 2

Question: 3 – In the following APs, find the missing terms in the boxes

(i) 2, (...), 26

Solution: We know that in AP, middle term is average of the other two terms

Thus, middle term `= (2 + 26)/(2) = (28)/(2) = 14`

Thus, above AP can be written as 2, 14, 26

(ii) (...), 13, (...), 3

Solution: Here, the middle term between 13 and 3 will be;

`(13 + 3)/(2) = (16)/(2) = 8`

Now, `a_4 – a_3 = 3 – 8 = - 5`

`a_3 – a_2 = 8 – 13 = - 5`

Thus, `a_2 – a_1 = - 5`

Or, `13 – a_1 = - 5`

Or, `a_1 = 13 + 5 = 18`

Thus, above AP can be written as 18, 13, 8, 3

(iii) 5, (...), (...), 9½

Solution: We have, a = 5 and a₄ = 9½

Now common difference can be calculated as follows:

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

Or, `91/2=5+3d`

Or, `3d=(19)/(2)-5=9/2`

Or, `d=9/2xx1/3=3/2`

Using common difference, 2nd and 3rd terms can be calculated as follows:

`a_2=a+d`

`=5+3/2=(13)/(2)`

`a_3=a+2d`

`=5+2xx3/2=8`

Thus, the AP can be written as; 5, 13/2, 8, 19/2

(iv) – 4, (...), (...), (...), (...), 6

Solution: Here, a = - 4 and a6 = 6

Common difference can be calculated as follows:

`a_6 = a + 5d`

Or, `6 = -4 + 5d`

Or, `5d = 6 + 4 = 10`

Or, `d = 2`

The second, third, fourth and fifth terms of this AP can be calculated as follows:

`a_2 = a + d = - 4 + 2 = - 2`

`a_3 = a + 2d = - 4 + 4 = 0`

`a_4 = a + 3d = - 4 + 6 = 2`

`a_5 = a + 4d = - 4 + 8 = 4`

Thus, the given AP can be written as: - 4, - 2, 0, 2, 4, 6

(v) (...), 38, (...), (...), (...), - 22

Solution: Let us take 38 as the first term and – 22 as the 5th term

Using this, common difference can be calculated as follows:

`a_5 = a + 4d`

Or, `- 22 = 38 + 4d`

Or, `4d = - 22 – 38 = - 60`

Or, `d = - 15`

If 38 is the second term, then first term can be calculated as follows:

`a = a_2 – d = 38 + 15 = 53`

Third, fourth and fifth terms can be calculated as follows:

`a_3 = a + 2d = 53 + 2(- 15) = 53 – 30 = 23`

`a_4 = a + 3d = 53 – 45 = 8`

`a_5 = a + 4d = 53 – 60 = - 7`

So, the AP can be written as: 53, 38, 23, 8, - 7, - 22