Find the sum of the first 10 terms.

8,20,32,44 ...

8,20,32,44,56,68,80,92,104,116,128

Step-by-step explanation

Just keep adding 12

Answer: 620

Step-by-step explanation:

This is a progression but we need to find out whether it is Arithmetic Progression or an exponential or geometric progression.

From the quest, 8, 20, 32, 44, this is not an exponential function but an arithmetic progression because, when you subtract the first time from the second term, the common difference is 12.

20 - 8 = 12, so this is an arithmetic progression as said earlier. Now to find the sum of the first ten terms of the series, we apply the formula which is

S10 = n/2{ (2a + (n - 1) d}, where n = 10, and a = 8 the first term and d the common difference = 12 ... Now substitute for the values

S10 = 10/2{ (2 x 8 + (10 - 1) 12}

= 5 (16 + 9 x 12)

= 5 (16 + 108)

= 5 (124)

= 620.

Therefore, the sum of the 10 terms of the series is 620.

But we can also solve it using other means. To solve this we first find the last term which is the 10th term using this formula

T10 = a + (n - 1) d

= 8 + (10 - 1) 12

= 8 + 9 x 12

= 8 + 108

= 116.

Now with this we could put this in the equation below.

S10 = n/2 (a + L) where L is the last term calculated above. Now we substitute to get the sum.

S10 = 10/2 (8 + 116)

= 5 (8 + 116)

= 5 (124)

= 620.

I could have explained how we arrived at the formula above but the time could not permit, I will discuss that in the next lecture if I come across such question.