Question

In an AP, an = 4, d = 2, Sn = -14 find n and a.

Answer 1

$a_{n}$=a+(n-1)d=4
or, a+(n-1)2=4
or, a+2n-2=4
or, a+2n=4+2
or, a+2n=6
or, a=6-2n
$S_{n}$=n/2[2a+(n-1)d]=-14
or, n/2[2a+(n-1)2]=-14
or, an+n(n-1)=-14
or, (6-2n)n+n²-n=-14
or, 6n-2n²+n²-n=-14
or, -n²+5n=-14
or, n²-5n=14
or, n²-5n-14=0
or, n²-7n+2n-14=0
or, n(n-7)+2(n-7)=0
or, (n-7)(n+2)=0
either, n-7=0
or, n=7
or, n+2=0
or, n=-2
∵, n can not be negative ;
∴, n=7
∴, a=6-(2×7)
=6-14
=-8
∴, n=7 and a=-8 Ans.

Answer 2

Answer:

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Step-by-step explanation:

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