From Arithmetic Progression, Sums of n terms using formulae: $T(n)=a+(n-1)d, S(n)=n/2[2a+(n-1)d]$

Question

How to find the below questions answer? I have tried but can't find anything? I think it's too much tough! Please anybody can help me?

Find the sums indicated below:

3+6+9+...+300

Answer 1

Hint: We have $T(n)=a+(n-1)d$ and $S(n)=\frac{n}{2}[2a+(n-1)d]$.

Let $T(1)=3$ and $T(n)=300$.

(1) What is $a$?

(2) Find $d$.

(3) Find $n$.

(4) Find $S(n)$

Answer 2

1. $a$ is the first term of the series, so this is equal to $3$.
2. $d$ is the difference between the first and next term. This is also equal to $3$.
3. $n$ is the number of terms in the sequence from 3 to 300. Since $T(n) = 300$, $a = 3$ and $d = 3$, this gives us $300 = 3 + (n-1)3$, or $n = 100.$
4. For $S(n)$, plug in $a$, $n$ and $d$. This gives us $15150.$

In a neater form, this is $\sum_{n=1}^{100} 3n = 15150$, but we can move the constant ($3$) outside the summation sign note that $3 \times \sum_{n=1}^{100} n = 3 \times 5050 = 15150.$