Question

Question 13 (QID: 146650647)
Instructions:Select the ONE correct answer from the given options.
1+4+7+ ... +(3n - 2) =
(1) 1/2n(3n-1)
(2) 1/2n(4n-1)
(3) 1/2n(2n-1)
(a) 1/2n(n-1)

Answer 1

Answer:

Option A

Step-by-step explanation:

This is an Arithmetic progression with common difference (d) as 3 ( 4-1 = 3 , 7-4 = 3, etc )

Now , in an Arithmetic progression , $n^{th}$ term is given by :

$a_{n}$ = a + (n-1)d

Where , $a_{n}$ = $n^{th}$ term

             n = number of terms

             a = first term

             d = common difference .

Let's say that (3n-2) is the $x^{th}$ term of the series .

Plugging them into the formula , we get :

(3n-2) = 1 + (x-1) 3

3n-2 = 1 + 3x - 3

     3x = 3n   ===> x = n .

Now , to find the Sum of first n terms , the formula is :

$S_{n}$ = n/2 *  ( 2a + (n-1)d )

$S_{n}$ = n/2 * ( 2(1) + (n-1)3 )

$S_{n}$ = n/2 * ( 3n -1 )

$S_{n}$ = 1/2 *  ( n(3n-1) )

Which is Option A .

Hope this answer helps you .