Question

how many terms of the AP - 15 - 13 - 11 are needed to make the sum -55

Answer 1

t1=-15=a
common difference is -13-(-15)=2=d
Sn=-55

$sn ={ n\%2(2a + (n - 1)d}$
-55=[n/2(2*-15+(n-1)2]
-55=[n/2(-30+2n-2]
-110=n(2n-32)
-110=2n²-32n
2n²-32n+110=0
n²-16n+55=0
n²-11n-5n+55=0
n(n-11)-5(n-11)=0
(n-11) (n-5)
n=11 or n=5
ans is 5

Answer 2

Answer:

5

Explanation:

t1=-15=a

common difference is -13-(-15)=2=d

Sn=-55

-55=[n/2(2*-15+(n-1)2]

-55=[n/2(-30+2n-2]

-110=n(2n-32)

-110=2n²-32n

2n²-32n+110=0

n²-16n+55=0

n²-11n-5n+55=0

n(n-11)-5(n-11)=0

(n-11) (n-5)

n=11 or n=5

ans is 5