how many terms of the a.p. 120,114,108,....be taken so that their sum is 1260?
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120 , 114 , 108 , ......
first term of AP is 120
and common difference is -6
Use formula ,
Sn=n/2 {2a +(n-1) d }
1260 =n/2 {2.120+(n-1)(-6)}
=n {120 -3n+3}
3n^2-123n+1260 =0
n^2 -41n+420=0
use quadratic formula ,
n={41+_root (41^2-4x420)}/2
=(41 +_1)/2
=20 , 21
here you see ,
n=20, and 21 two number possible .
because-----
T21=120+(21-1)(-6)=0
T20=120-19x6=6
here you can see that 21th term is zero so sum is independent of 21th term .
hence ,
n =20 and 21 both value possible
first term of AP is 120
and common difference is -6
Use formula ,
Sn=n/2 {2a +(n-1) d }
1260 =n/2 {2.120+(n-1)(-6)}
=n {120 -3n+3}
3n^2-123n+1260 =0
n^2 -41n+420=0
use quadratic formula ,
n={41+_root (41^2-4x420)}/2
=(41 +_1)/2
=20 , 21
here you see ,
n=20, and 21 two number possible .
because-----
T21=120+(21-1)(-6)=0
T20=120-19x6=6
here you can see that 21th term is zero so sum is independent of 21th term .
hence ,
n =20 and 21 both value possible
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