Question

Let there be an A.P. with first term 'a', common difference 'd'. If an denotes its nth term and Sn
the sum of first n terms, find n and a, if an = 4, d = 2 and Sn = -14.​

Answer 1

Answer:

Since an=4

a+(n-1)2=4 given d=2

a+2n=6 ...........(1 )

Sn=-14

n/2{2a+(n-1)2}=-14

n{a+n-1}=-14

n(6-2n+n-1)=-14 from 1 a=6-2n

n(5-n)=-14

5n-n^2=-14

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 n=-2

Using this value in equation (1)

a=-8 & a=10

Answer 2

Step-by-step explanation:

Given that,

first term of the A.P.= a

common difference = d=2

sum of the nth term Sn= -14

an= 4

Since we know that,

nth term of the A.P(an)= a+(n-1)d

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

=>4 = a + 2n - 2

=> 6 = a + 2n ------------(i)

Again,

Sum of it's n terms Sn

$sn = \frac{n}{2}(a + l) \\ = > - 14 = \frac{n}{2}(a + 4) \\ = > - 28 = n(a + 4) \\ = > - \frac{28}{a + 4} = n \: \: \: \: \: \: \: (ii)$

Putting the value of n in equation (i), we get,

$6 = a + ( - \frac{ 2 \times 28}{a + 4} ) \\ = > 6 = \frac{ {a}^{2} + 4a - 56 }{a + 4} \\ = > 6a + 24 = {a}^{2} + 4a - 56 \\ = > {a}^{2} - 2a - 80 = 0 \\ = > {a}^{2} - 10a + 8a - 80 = 0 \\ = > a(a - 10) + 8(a - 10) = 0 \\ = > (a - 10)(a + 8) = 0$

Either , a = 10 or a = - 8

Since, Sum is negative value ,

Therefore, a = - 8

Now , putting the value in equation (i)

6 = -8 +2n

=> 14 = 2n

=> n = 7

Therefore, value of first term of the A.P. is -8 and number of term = 7

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