Question

if the first term of an AP is -24 and common different is -4, then find the sum of its first 12 term​

Answer 1

Step-by-step explanation:

sum = n/2 [first term + last term]

last term:

first term + difference(n-1),   here n = 12

-24 + (-4)(12-1)

=> -68

sum:

=> 12/2 [-24 + (-68)]

=> 6 (-92)

=> -552

Thus the sum of the first 12 terms is -552

Answer 2

Given:-

a = t1 = -24

common difference = d = -4

To Find:-

Sum of its 12th term. i.e S12 = ?

Solution:-

a = -24 , d = -4, n = 12

Sn = n/2[2a + (n - 1)d]

s12 = 12/2 [ 2 × (-24) + (12 - 1) -4]

s12 = 6[ -48 + (11) × -4]

s12 = 6( -48 - 44)

s12 = 6 ( -92)

s12 = -552

Answer:-

Sum of 12 term is -552

More information:-

For the given Arithmetic Progression, if first term is a and common diffrence is d then

=> tn = [a + (n - 1)d] and

=> sn = n/2 [2a + (n - 1)d]

In the A.P. a, a + d, a + 2d, a + 3d, . . . a +(n - 1)d

First term = t1 = a and nth term is [a +(n - 1)d ]

=> Sn = n/2 [ t1 + tn]