Question

please help class 10th questions

Answer 1

Step-by-step explanation:

n = 7 and a = - 8

since, an= a+(n-1)d

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

=> 4= a + 2n -2

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

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

Again

Sn= (n/2){a+an}

-14=(n/2)(a+4)

=>-28 = n(a+4) -----------(iii)

Putting the value of a in equation (iii),

-28= n(6-2n+4)

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

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

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

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

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

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

Either ,. or

n-7= 0. n+2=0

=>n= 7. =>n = -2

Since number of terms cannot be negative.

Therefore, n= 7

Putting the value of n in equation (ii)

a= 6 -14

=> a = -8

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

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Answer 2

Answer:

let a be the first term of the given A.P.

An = a + ( n-1 ) d

4 = a+(n-1) 2

4- 2n + 2 = a

6- 2n =a

Step-by-step explanation:

Sn = n\2 [ a + l ]

-14= n\2 [ a + 4]

-14 x 2 =n( a+4)-----(2)

Putting a = 6 - 2 in ----(2)

-28 = n( 6 - 2 + 4)

-28 = (10 -2n)

-28 = 10n - 2n^2

2n^2-10n - 28 = 0

2(n^2- 5n -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

let n-7=0

=>n -7               when n + 2 =0

=>n=7                    => n= -2

Putting n= 7 in (1)

6 - 2 (7) =a

6 - 14 =a

=> -8= a

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