Arithmetic Progression

NCERT Exercise 5.3

Part 1

Question: 1 – Find the sum of the following APs:

(i) 2, 7, 1, 2, ………………….. to 10 terms

Answer: Here, a = 2, d = 5 and n = 10

Sum of n terms can be given as follows:

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

Or, `S_(10)=(10)/(2)[2xx2+(10-1)5]`

`=5(4+45)=5xx49=245`

Thus, sum of the 10 terms of given AP (S_n)=245

(ii) – 37, – 33, – 29, ………………..to 12 terms

Answer: Here, a = - 37, d = 4 and n = 12

Sum of n terms can be given as follows:

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

Or, `S_12=(12)/(2)[2(-37)+11xx4]`

`=6(-74+44)=6xx30=-180`

Thus, sum of the 12 terms of given AP (S_n)= – 180

(iii) 0.6, 1.7, 2.8, ……………… to 100 terms

Answer: Here, a = 0.6, d = 1.1 and n = 100

Sum of n terms can be given as follows:

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

`S_(100)=(100)/(2)[2xx0.6+99xx1.1]`

`=50(1.2+108.9)`

`=50xx110.1=5505`

Thus, sum of the 100 term of given AP (S_n)= 5505

(iv) 1/15, 1/12, 1/10, ……… to 11 terms.

Answer: Here, a = 1/15, n = 11

`d=(1)/(12)-(1)/(15)`

`=(5-4)/(60)=(1)/(60)`

Sum of n terms can be given as follows:

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

`=(11)/(2)[2xx(1)/(15)+10xx(1)/(60)]`

`=(11)/(2)((2)/(15)+1/6)`

`=(11)/(2)((4+5)/(30))`

`=(11)/(2)xx(9)/(3)=(23)/(20)=1(13)/(20)`