Question

Find the sum of the following arithmetic progressions:

41, 36, 31, ... to 12 terms

Answer 1

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

41, 36, 31, ... to 12 terms

Common difference of the A.P. (d) = ${\displaystyle a_2-a_1}$

= 36 - 41

= -5

Number of terms (n) = 12

The first term for the given A.P. (a) = 41

So, using the formula we get,

${\displaystyle S_{12}=\frac{12}{2}[2\left(41\right)+(12-1)(-5)]}$ 

= (6)[82 + (11)(-5)]

= (6)[82 - 55]

= (6)[27]

= 162

Therefore the sum of first 12 terms for the given A.P is 162