Question

In A.P 6, 12, 18 find S40 ​

Answer 1

Given sequence of AP is

6, 12, 18, 24, ....................

Here,

first term, a = 6

Common difference, d = 12 - 6 = 6

So, S(40) = sum upto 40 terms

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

Where n = no. Of terms

Now, S(40) = 40/2[2*6 + (40 - 1)6]

→ S(40) = 20[12 + 39*6]

→ S(40) = 20(12 + 234)

→ S(40) = 20(246)

→ S(40) = 4920 ______________(Ans.)

Answer 2

Answer:

4920

Step-by-step explanation:

S40 = 20[12+39(6)]

       = 20[246]

       = 4920