Question

An= 4 , d= 2 , sn= -14 .. find n and a

Answer 1

Answer:

n = 7 , a = -8

Step-by-step explanation:

Formula used = an= a+(n-1)d

an =a+(n-1)×d

=> 4 = a + ( n - 1 ) × 2

=> 4 = a + 2n -2

=> 4 + 2 = a + 2n

=> a + 2n = 6 .

=> a = 6 - 2x

=> a = 6 - 2x

=> sn = $\dfrac{n}{2}$ [ 2a + ( n - 1 ) × d ]

=> -14 = $\dfrac{n}{2}$ [2a + ( n - 1 ) × 2 ]

=> -28 = n ( 2a + 2n - 2)

=> - 28 = n [ 2 ( 6 - 2n ) + 2n - 2 ]

=> -28 = n ( 10 - 2n )

=> - 28 = 10 x - 2n

=> - 2n^2 +10 x +28 = 0 .

=> n ^2 - 5n - 14 = 0.

=> n^2- 7 n + 2 n -14 = 0

=>n(n - 7 ) + 2( n- 7) = 0

=> ( n - 7) (n + 2) = 0

=> x - 7 = 0

• x = +7

=> x + 2 = 0

• x = -2

so , n = 7 as 2 is in negitive .

• a = 6 - 2n .
• a = 6 - 2 × 7
• a = 6 - 14
• a = -8 .

.....Hope this will help u .....

Answer 2

Answer:

a= -8 and n=7

Step-by-step explanation:

an=a+(n-1)d

=> 4=a+(n-1)2

=> 4=a+2n-2

=> 4+2=a+2n

=> 6=a+2n

=> a= 6-2n-------(1)

sn=n/2[2a+(n-1)d

=> -14=n/2[2a+(n-1)2]

=> -14*2=n(2a+2n-2)

=> -28=n(2a+2n-2)

=> -28=n[2(6-2n)+2n-2]...............from equation (1)

=> -28=n(12-4n+2n-2)

=> -28=n(-2n+10)

=> -28= -2n^2+10n

=> 2n^2-10n-28=0

=> 2(n^2-10n-14)=0

=> n^2-5n-14=0

=> n^2-7n+2n-14=0

=> n(n-7)+2(n-7)=0

=> (n-7)(n+2)=0

=> n-7=0

=> n=7

and... n+2=0

n= -2

n=7 and -2

n will not be negative

so, n=7

put the value n in equation.... (1)

a=6-2n

a=6-2*7

a=6-14

a= -8

hence, a= -8 and n=7 answer