Question

How many terms of the ap 27 , 24 , 21 ... Should be taken so that their sum is zero?

Answer 1

Answer:

Here's your answer

Given

a= 27 , d = 24-27=-3

Sum =0

$s = \frac{n}{2} (2a + (n - 1)d)$

$s = \frac{n}{2} (54 - </strong><strong>3</strong><strong>n + </strong><strong>3</strong><strong>) \\ \\ 0 = n(56 - 2n) \\ \\ 0 = 5</strong><strong>7</strong><strong>n - </strong><strong>3</strong><strong> {n}^{2} \\ \\ </strong><strong>3</strong><strong> {n}^{2} - 5</strong><strong>7</strong><strong>n = 0 \\ \\ </strong><strong>3</strong><strong>n(n - </strong><strong>1</strong><strong>9</strong><strong>) = 0$

n=19....

Therefore number of terms are 19......

Hope it helps ✌✌.....

Answer 2

GIVEN :

AP = 27, 24, 21....

The first term = a = 27

Common Difference(d) = -3

Sum of terms = Sn = 0

We know that,

Sum of terms in an AP = n/2 [2a + (n - 1)d]

0 = n/2 [ 2(27) + (n - 1)-3 ]

0 = n/2 [ 54 - 3n + 3 ]

0 = n/2 [ 57 - 3n ]

0 × 2/n = 57 - 3n

=> 57 - 3n = 0

=> 3n = 57

=> n = 57/3

=> n = 19

Therefore, 19 terms should be taken.