Saved Bookmarks
| 1. |
③Findthesum of first8 multiples of. |
|
Answer» =sum of first 8 multiples of =3=3+6+9+12+15+18+21+24=3(1+2+3+4+5+6+7+8)=summation of natural no's==n(n+1)/2=8(8+1)/2=8(9)/2=36sum of first 8multiples of 3=3(36)=108 3+6+9+12+15+18+21+24=108 It will make a series having first term, a=3 and common difference, d=3Then we will apply formula of A.PSum of first 8 multiples of A.P={2a+(n-1) d}×n/2={2×3+7×3}×8/2=27×4=108 The first-8 multiples of 3 are 3,6,9,12,15,18,21 and 24these numbers are now in Arithmetic progressionsofirst term, a=3common difference,d=6-3=3number of terms,n=8Therefore Sum of AP=n/2[2a+(n-1)d]=8/2[2(3)+(8-1)3]=4[6+(7)3]=4(6+21)=4x27=108 a=3, d= 3sum = n/2{2a+(n-1)d} =8/2{2×3+(8-1)×3} = 4(6+21) = 27×4 = 108 108 is the best answer |
|