Arithemtic Progression NCERT Exercise 5.1 part two Class Ten Mathematics

Arithmetic Progression

Exercise 5.1 Part 2

Question: 2 – Write first four terms of AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

Solution: Here, first term a1 = 10 and common difference, d = 10

Hence, second term `a_2 = a_1 + d = 10 + 10 = 20`

Third term `a_3 = a_1 + 2d = 10 + 2 xx 10 = 30`

Fourth term `a_4 = a_1 + 3d = 10 + 30 = 40`

Thus, first four terms of the AP will be 10, 20, 30, 40, ……


(ii) a = - 2, d = 0

Solution: First term a = 1

Common difference = 0

Thus, first four terms of given AP will be

`a_1 = - 2`, `a_2 = - 2`, `a_3 = - 2` and `a_4 = - 2`

(iii) `a = 4`, `d = - 3`

Solution: Here, first term a1 = 4 and common difference d = - 3

We know that `a_n = a + (n – 1)d`, where n = number of terms

Thus, second term `a_2 = a + (2 – 1)d`

Or,`a_2 = 4 + (2-1)xx (-3)`

`= 4 - 3 = 1`

Similarly, third term `a_3 = a + (3 – 1)d`

`a_3 = 4 + (3-1) xx (-3)`

`= 4 - 6 = -2`

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

`a_4 = 4 + (4 - 1) xx( -3)`

`= 4 - 9 = -5`

Thus, the first four terms of given AP are; 4, 1, - 2, - 5


(iv) a = - 1, d = ½

Solution: Solution: We have, first term = - 1 and d = ½

`a_2=a+(2-1)d`
`=-1+1/2=-1/2`

`a_3=a+2d`
`=-1+2xx1/2=0`

`a_4=a+3d`
`=-1+3xx1/2=1/2`

Thus, the four terms are; - 1, -1/2, 0 and ½

(v) a = - 1.25, d = - 0.25

Solution: We have; first terms = - 1.25 and d = - 0.25

`a_2= a + d`

`= -1.25 - 0.25 = -1.5`

`a_3 = a + 2d`

`= -1.25 + 2 xx (-0.25)`

`= -1.25 - 0.5 = -1.75`

`a_4 = a + 3d`

`= 1.25 + 3 × (-0.25) = -2.25`

Thus, the first four terms are; - 1.25, -1.5, -1.75 and – 2.25


Question: 3 – For the following APs, write the first term and common difference,

(i) 3, 1, -1, - 3, …….

Solution: Here, first term a = 3

Common difference can be calculated as follows:

`a_4 – a_3 = - 3 – ( -1) = - 3 + 1 = - 2`

`a_3 – a_2 = - 1 – 1 = - 2`

`a_2 – a_1 = 1 – 3 = - 2`

Here, `a(k+1) – a_k = - 2` for all values of k

Hence, first term = 3 and common difference = - 2

(ii) – 5, - 1, 3, 7

Solution: `a_4 – a_3 = 7 – 3 = 4`

`a_3 – a_2 = 3 – (-1) = 3 + 1 = 4`

`a_2 – a_1 = - 1 – (-5) = - 1 + 5 = 5`

Here, `a_(k+1) – a_k = - 2` for all values of k

Hence, first term = - 5 and common difference = 4

Question (iii): `1/3`, `5/3`, `9/3`, `(13)/(3)`, ------

Solution:

`a_4-a_3=(13)/(3)-9/3=4/3`

`a_3-a_2=9/3-5/3=4/3`

`a_2-a_1=5/3-1/3=4/3`

Here, `a_(k+1) – a_k = - 2` for all values of k

Hence, first term = 1/3 and common difference = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, …….

Solution: `a_4 – a_3 = 3.9 – 2.8 = 1.1`

`a_3 – a_2 = 2.8 – 1.7 = 1.1`

`a_2 – a_1 = 1.7 – 0.6 = 1.1`

Here, `a_(k+1) – a_k = - 2` for all values of k

Hence, first term = 0.6 and common difference = 1.1



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