NCERT Solutions
Class 10 Maths
Arithmetic Progressions

Ex. 5.3 Q.1
Find the sum of the following APs:
(i) 2, 7, 12, . . ., to 10 terms. (ii) -37, -33, -29, . . ., to 12 terms.
(iii) 0.6, 1.7, 2.8, . . ., to 100 terms. (iv) 1/15,1/12,1/10,.....,to 11 terms.
Sum of n terms of Arithmetic progression (AP) is Sum = (n/2)*{2a + (n - 1)*d} where n = Number of terms in AP
a = First term of AP d = Common difference of AP
- Given Arithmetic Series is:
2,7,10,.....,to 10 terms
Here a =2, n = 10, d = 7-2 = 5
Sum = (10/2)*{2*2 + (10-1)*5}
= 5*(4 + 9*5)
= 5*(4 + 45)
= 5*49
= 245
=> Sum = 245
- Given Arithmetic Series is:
-37,-33,-29,.....,to 12 terms
Here a =-37, n = 12, d = -33-(-37) = -33 + 37 = 4
Sum = (12/2)*{2*(-37) + (12-1)*4}
= 6*(-74 + 11*4)
= 6*(-74 + 44)
= 6*(-30)
= - 180
=> Sum = - 180
- Given Arithmetic Series is:
0.6,1.7,2.8,.....,to 100 terms
Here a =0.6, n = 100, d = 1.7 - 0.6 = 1.1
Sum = (100/2)*{2*0.6 + (100-1)*1.1}
= 50*(1.2 + 99*1.1)
= 50*(1.2 + 108.9)
= 50*110.1
= 5505.0
=> Sum = 5505.0
- Given Arithmetic Series is:
1/15,1/12,1/10,.....,to 11 terms
Here a =1/15, n = 11, d = 1/12 - 1/15 = 1/60
Sum = (11/2)*{2*(1/15) + (11-1)*(1/60)}
= (11/2)*(2/15 + 10/60)
= (11/2)*(18/60)
= (11*18)/(2*60)
= (11*3)/(2*10) (when 18 and 60 is divided by 6) = 33/20
=> Sum = 33/20