Question

What is the sum of the arithmetic series 18 t=1 (3t-4)

Answer 1

Consider the arithmetic series $_{t=1}\Sigma^{t=18} (3t-4)$

Let t=1 in the given series, we get

first term = $a_{1}$ = 3-4 = -1.

Let t=2 in the given series, we get

second term = $a_{2}$ = $(3 \times 2)-4 = 2$

Let t=3 in the given series, we get

third term = $a_{3} = (3 \times 3)-4=5$

Now, let t=18 in the given series, we get

last term = l = $l = (3 \times 18)-4 = 50$

We get the series as

-1, 2, 5,..... 50

Sum = $\frac{n}{2}(a+l)$

= $\frac{18}{2}(-1+50)$

= $= 9 \times 49$

= 441

Therefore, the sum of the given arithmetic series is 441.

Answer 2

The given arithmetic series is

$\sum_{t=1}^{18} (3t-4)$

When t =1, we will get the first term,a, which is

$3(1)-4 =3-4=-1$

When t =18, we will get the last term,l , which is

$3(18)-4 = 50$

Now we use the formula of sum of n terms which is

$S = \frac{t}{2} (a+l)$

Substituting the values of t,a and l, we will get

$S = \frac{18}{2} (-1+50) = 9*49 =441$