Saved Bookmarks
| 1. |
How many terns cf the AP 3, 5,7,9,... must be added to get the sum 120? |
|
Answer» we have to find n Sn = 120a= 3d= 2Sn = n/2+(2a+(n-1)d120=n/2(2*3+(n-1)2)240= n(6+2n-2)240= 6n+2n^2-2n240= 4n+2n^2120= 2n+n^2n^2+2n-120=0n^2+(12-10)n-120=0n^2+12n-10n-120=0n(n+12)-10(n+12)=0n=10and n= -12number of terms is 10 |
|