Q.Find the sum of the first 20 terms of the Arithmetic Progression $4, 7, 10, \ldots$

Use the sum formula $S_n=\frac{n}{2}[2a+(n-1)d]$ with $a=4,d=3,n=20$.

Here $a=4$, $d=7-4=3$ (checked: $10-7=3$ — constant), $n=20$.

Apply $S_n=\dfrac{n}{2}[2a+(n-1)d]$: $S_{20}=\dfrac{20}{2}[2(4)+(20-1)(3)]=10[8+57]=10(65)=650$. …

Alternative — pairing method directly: pairing the 1st and 20th term ($4+61=65$), the 2nd and 19th term (also $65$, since each pair moves one step in from opposite e …

A common error is using $n$ instead of $(n-1)$ inside the bracket $[2a+(n-1)d]$, giving a wrong extra $d$: here that mistake would compute $[8+20 …