Arithmetic Progression NCERT Exercise 5.3 part two Class Ten Mathematics

Arithmetic Progression

Exercise 5.3 Part 2

Question: 2 – Find the sums given below:

(i) `7 + 10.5 + 14 + ….. + 84`

Solution: Here, a = 7, d = 3.5 and last term = 84

Number of terms can be calculated as follows;

`a_n =a + (n-1)d`

Or, `84 = 7 + (n-1)3.5`

Or, `(n-1)3.5 = 84-7`

Or, `n - 1 = 77÷3.5 = 22`

Or, `n = 23`


Sum of n terms can be given as follows:

`S=n/2[2a+(n-1)d]`

`=(23)/(2)[2xx7+22xx3.5]`

`=(23)/(2)(14+77)`

`=(23)/(2)xx91=(2093)/(2)`

`=1046(1)/(2)`

(ii) 34 + 32 + 30 + …. + 10

Solution: Here, a = 34, d = - 2 and last term = 10

Number of terms can be calculated as follows:

`a_n = a + (n – 1)d`

Or, `10 = 34 + (n – 1)(- 2)`

Or, `10 = 34 – (n – 1)(2)`

Or, `(n – 1)2 = 34 – 10 = 24`

Or, `n – 1 = 12`

Or, `n = 13`


Sum of n terms can be given as follows:

`S=n/2[2a+(n-1)d]`

`=(13)/(2)[2xx34-12(-2)]`

`=(13)/(2)(68-24)`

`=(13)/(2)xx44=286`

Thus, sum of given AP = 286


(iii) – 5 + (-8) + (- 11) + …… + (- 230)

Solution: Here, a = - 5, d = - 3 and last term = - 230

Number of terms can be calculated as follows:

`a_n = a + (n – 1)d`

Or, `- 230 = - 5 + (n – 1)( - 3)`

Or, `- 230 = - 5 – (n – 1)3`

Or, `(n – 1)3 = - 5 + 230 = 225`

Or, `n – 1 = 75`

Or, `n = 76`

Sum of n terms can be given as follows:

`S=n/2[2a+(n-1)d]`

`=(76)/(2)[2(-5)+75(-3)]`

`=38(-10-225)`

`=38(-235)=-8930`

Thus, sum of given AP = - 8930



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