Question

How many terms of AP 10,8,6 ....must be taken to give the sum of -126

Answer 1

Answer:

18 terms will be required

Step-by-step explanation:

sum of AP : Sn = n/2 [2a+(n-1)d]

a=10, a2=8

d = a2-a = 8-10 = -2

Sn = -126

-126 = n/2 [20+(n-1)(-2)] = n/2 (20-2n+2)

from rhs taking 2 common

-126 = n(10-n+1) = 11n-n^2

forming quadratic equation in n :

n^2 - 11n - 126 = 0

factors of (-126) to form 11 >>> (-18) and (+7)

equation : n^2 -18n + 7n - 126 = 0

factorising : (n+7)(n-18) = 0

this gives n = -7 or n = +18

since n = no. of terms cannot be negative

n=18

Answer 2

$\huge\bold{Answer:-}$

Given:-

The Ap is 10 , 8 , 6 , .....

The sum of the terns of Ap is -126

To Find:-

The number of terms so that the sum becomes -126

Solution:-

Let, the first term of the ap be a is 10

The common difference be d = ( 8 - 10) = -2

So, to find the number of terms so that the sum becomes -126 , we have to do with the formulla-

Sn = $\dfrac{n}{2}$ × {2a + ( n -1 ) d}.

where,

s= sum of number of terms .

n = number of terms

a = first term of the Ap

d = common difference .

Now, .

Sn = $\dfrac{n}{2}$ × {2a + ( n -1 ) d}.

-126 = $\dfrac{n}{2}$ × {2 × 10 + ( n -1 ) (-2)}.

-126 × 2 = n × { 20 + ( -2n + 2 ) }

-252 = n ( 20 - 2n + 2)

-252 = 20n - 2n² + 2n

22n - 2n² + 252 = 0

-2n² + 22n + 252 = 0

Taking -2 as common ,

-2( n² - 11n - 126 ) = 0

n² - 11n - 126 = 0

n² - ( 18 - 7 )n - 126 = 0

n² - 18n + 7n - 126 = 0

n ( n - 18 ) + 7 ( n - 18 ) = 0

( n - 18 ) ( n + 7 ) = 0

either,

( n - 18 ) = 0

n = 18

or, .

( n + 7 ) = 0

n = -7

As, no of terms cannot be negative so , 18 no of terms should be needed so that the sum makes -126.