Question

Let $V_r$ denote the sum of the first $r$ terms of an A.P whose first term is $r$ and common difference is $(2r-1)$.Let
$T_r=V_{r+1}-V_r-2$ is always

• A. an odd number
• B. an even number
• C. a prime number
• D. a composite number

Answer 1

The correct option is D a composite number

We have sum of $n$ terms $=\frac{n}{2}(2a+(n-1\left)d\right)$ where $a$ is the first term, $n$ is the number of terms and $d$ is the common difference in an A.P.
From the passage $n=r,$ $a=2r$ and $d=(2r-1)$
$\therefore V_r=\frac{r}{2}[2r+(r-1\left)\right(2r-1\left)\right]$
$\Rightarrow \frac{r}{2}[2r+2r^2-3r+1]=\frac{r}{2}[2r^2-r+1]$
Thus $V_r=\frac{1}{2}[2r^3-r^2+r]=r^3-\frac{r^2}{2}+\frac{r}{2}$
Now $T_r=V_{r+1}-V_r-2$
From above $V_r=r^3-\frac{r^2}{2}+\frac{r}{2}$
$V_{r+1}={(r+1)}^3-\frac{{(r+1)}^2}{2}+\frac{(r+1)}{2}$
We have $T_r=V_{r+1}-V_r-2$
$T_r=r^3-\frac{r^2}{2}+\frac{r}{2}-({(r+1)}^3-\frac{{(r+1)}^2}{2}+\frac{(r+1)}{2})-2$
On simplifying, we get

$T_r={(r+1)}^3-r^3-\frac{1}{2}({(r+1)}^2-r^2)+\frac{1}{2}(r+1-r)-2$
$\Rightarrow T_r=(r+1-r)({(r+1)}^2+r(r+1)+r^2)+\frac{1}{2}-2$
On simplifying, we get
$T_r=r^2+1+2r+r^2+r+r^2+\frac{1}{2}(-2r-1+1)-2$
$T_r=3r^2+2r-1=(3r-1)(r+1)$
We have $T_1=(3-1)(1+1)=2.2$
$T_2=(3\times 2-1)(2+1)=5.3$
$T_3=(3\times 3-1)(3+1)=8.4$ which are in A.P
Thus,$T_n=(3n-1)(n+1)$
From the above sequence, we note that
Product of even number and an odd number is Even
Product of odd number and an odd number is odd
Product of even number and an even number is Even
and we see that every term is a composite number.
Hence, their sum is a composite number.
$\therefore T$ is a composite number.