Doubt by Ziya
Solution :
an=4
d=2
Sn=-14
We know,
an=a+(n-1)d
4=a+(n-1)2
4=a+2n-2
4+2=a+2n
6=a+2n
6-2n=a
an=a+(n-1)d
4=a+(n-1)2
4=a+2n-2
4+2=a+2n
6=a+2n
6-2n=a
a=6-2n — (1)
Sn=(n/2)[a+an]
-14=(n/2)[6-2n+4]
-14=(n/2)[10-2n]
-14=(n/2)×2[5-n]
-14=n(5-n)
-14=5n-n²
n²-5n-14=0
-14=(n/2)[6-2n+4]
-14=(n/2)[10-2n]
-14=(n/2)×2[5-n]
-14=n(5-n)
-14=5n-n²
n²-5n-14=0
n²-(7-2)n-14=0
n²-7n+2n-14=0
n²-7n+2n-14=0
n(n-7)+2(n-7)=0
(n+2)(n-7)=0
n+2=0
n=-2
But n can't be negative so rejected.
(n+2)(n-7)=0
n+2=0
n=-2
But n can't be negative so rejected.
(n-7)=0
n=7
Putting this value of n in equation (1)
n=7
Putting this value of n in equation (1)
a=6-2(7)
a=6-14
a=6-14
a=-8
Hence, n=7 & a=-8