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
Solution: 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 (Sn)=245
(ii) – 37, – 33, – 29, ………………..to 12 terms
Solution: 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 (Sn)= – 180
(iii) 0.6, 1.7, 2.8, ……………… to 100 terms
Solution: 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 (Sn)= 5505
(iv) 1/15, 1/12, 1/10, ……… to 11 terms.
Solution: 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)`