Question

Need help again asap please! Find the sum of the first 30 terms of the sequence below an=3n+2

Answer 1

$a_n=3n+2\\a_{n+1}=3(n+1)+2=3n+3+2=3n+5\\\\a_{n+1}-a_n=(3n+5)-(3n+2)=3n+5-3n-2=3=const.\\\\\text{It's an arithmetic sequence where}\\a_1=3(1)+2=3+2=5\ and\ d=3\\\\\text{The formula of a sum of the first n terms of an arithmetic sequence:}\\\\S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n\\\\\text{We have:}\\a_1=5,\ d=3\ \text{and}\ n=30\\\\\text{Substitute}\\\\S_{30}=\dfrac{2(5)+(30-1)(3)}{2}\cdot30=\dfrac{10+(29)(3)}{1}\cdot15=(10+87)(15)\\\\=(97)(15)=1455\\\\Answer:\ 1455$

Answer 2

$S_{30}$ = 1455

generate a few terms of the sequence using $a_{n}$ = 3n + 2

$a_{1}$ = ( 3 × 1) + 2 = 5

$a_{2}$ = (3 × 2) + 2 = 8

$a_{3}$ = (3 × 3 ) + 2 = 11

$a_{4}$ = (3 × 4 ) + 2 = 14

$a_{5}$ = ( 3 × 5 ) + 2 = 17

the terms are 5, 8, 11, 14, 17

these are the terms of an arithmetic sequence

sum to n terms is calculated using $S_{n}$ = $\frac{n}{2}$ [ 2a + (n-1)d]

where a is the first term and d the common difference

d = 8 - 5 = 11 - 8 = 14 - 11 = 3 and $a_{1}$ = 5

$S_{30}$ = $\frac{30}{2}$ [( 2 × 5) + (29 × 3) ]

= 15( 10 + 87) = 15 × 97 = 1455