Question

Find the sum of the first 12 terms of the sequence. 1, -4, -9, -14, . . . (2 points)

Answer 1

Answer:

sum of the first 12 terms of the sequence. 1, -4, -9, -14, . . = -318

Step-by-step explanation:

The given AP is,

1, -4, -9, -14, . . .

first term a = 1,

common difference d = -4 - 1 = -5

number of terms  n = 12

Equation:

Sum of n terms of AP, S₁₂ = n/2[2a + (n - 1)d]

S₁₂ = 12/2[(2*1) + (12 - 1)*(-5)]

S₁₂ = 6[2 + (11*(-5))]

S₁₂ = 6[2 - 55] = 6*(-53) = -318

Therefore sum of the first 12 terms of the sequence. 1, -4, -9, -14, . . = -318

Answer 2

Answer:

The sum of first 12 terms of the sequence. 1, -4, -9, -14, . .  is -318.

Step-by-step explanation:

Given sequence  1, -4, -9, -14, . . .

We have to find the sum of first 12 terms of the sequence. 1, -4, -9, -14, . .

Consider  the given sequence 1, -4, -9, -14, . . .

$a_1=1 ,a_2=-4,a_3=-9$

First calculate the common difference (d)

$a_2-a_1=-4-1=-5\\a_3-a_2=-9+4=-5$

Thus, common difference is -5

We know sum of terms in an Arithmetic progression is given by,

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

where n is number of terms ,

a = first term

d = common difference

Here, n = 12 , a= 1 , d = -5

$S_{12}=\frac{12}{2}(2(1)+(12-1)(-5))$

Solving , we get,

$S_{12}=6(2+11(-5))$

$S_{12}=6(2-55)$

$S_{12}=6(-53)$

$S_{12}=-318$

Thus, the sum of first 12 terms of the sequence. 1, -4, -9, -14, . .  is -318.