Question

How many terms of the Ap:-9, -6, -3,......must be taken so that their sum is 66

Answer 1

Step-by-step explanation:

a = - 9

d = - 9 - (-6) = -3

Sum of terms = n/2 (2a + (n - 1)d)

= 66 = n/2 (2 × -9 + (n - 1) × -3)

= 66 = n/2 (-18 - 3n + 3)

= 132 = - 18n - 3n² + 3n

= 132 = - 3n² - 15 n

= - 3n² - 15n - 132 = 0

Solve this quadratic equation

We should get n value as natural number

Answer 2

Here,

a=-6

d=-6-(-9)

=-6+9

=3

Let,

Sn=66

n/2{2*(-6)+(n-1)*3}=66

n{-12+3n-3}=66*2

n{-15+3n}=132

-15n+3n^2=132

3n^2-15n=132

n^2-5n-44=0

Just factorise it...