Question

Consider the arithmetic sequence 9, 15, 21
a) Write the algebraic form of this sequence.
b) Find the twenty fifth term of this sequence.
c) Find the sum of terms from twenty fifth to fiftieth of this sequence.
d) Can the sum of some terms of this sequence be 2015? Why?

Answer 1

Given arithmetic sequence is 9, 15, 21,.
First term = a = 9
Common difference = d =15 9 = 6
(A) For natural number n
nth term = $\left[\right(n-1\left)x6\right]+9=6n+3$ where $n=1,2,3,$
Hence algebraic form of the sequence $9,15,21,.$ is $x_n=6n+3$.
(B) nth term $=6n+3$
${25}^{th}$ term $=6\left(25\right)+3=150+3=153$
(C) Sum of first n terms is given by,
$S_n=\frac{1}{2}n[X_1+X_n]$
Sum of 24 terms is
$S_{24}=\frac{1}{2}n[X_1+X_{2}4]=\frac{25}{2}[9+147]=1950$
Sum of 50 terms is
$S_{50}=\frac{1}{2}n[X_1+X_{50}]=\frac{50}{2}[9+6(50)+3]=7800$
Sum of terms from twenty fifth to fiftieth
$=S_{50}-S_{24}$
$=7800-1950$
$=5850$
(D) Let sum of first n terms is 2015,
$S_n=\frac{1}{2}n[X_1+X_n]$
$2015=\frac{1}{2}n[9+6n+3]$
$2015=3n^2+6n$
$3n^2+6n-2015=0$
Solving above equation for n, the value of n is not a natural number.
Hence 2015 can not be the sum of some terms of this sequence.