Question

If n+1,2n+1,6n+3 are in GP then n

Answer 1

If $a,\ b,\ c$ are in GP then we have,

• $b^2=ac$

Here $n+1,\ 2n+1,\ 6n+3$ are in GP then we have,

$\longrightarrow (2n+1)^2=(n+1)(6n+3)$

$\longrightarrow 4n^2+4n+1=6n^2+9n+3$

$\longrightarrow2n^2+5n+2=0$

$\longrightarrow2n^2+4n+n+2=0$

$\longrightarrow2n(n+2)+(n+2)=0$

$\longrightarrow(n+2)(2n+1)=0$

$\Longrightarrow n=-2\quad OR\quad n=-\dfrac{1}{2}$

Taking $n=-2,$

• $n+1=-1$

• $2n+1=-3$

• $6n+3=-9$

Here -1, -3, -9 are in GP with first term -1 and common ratio 3.

Taking $n=-\dfrac{1}{2},$

• $n+1=\dfrac{1}{2}$

• $2n+1=0$

• $6n+3=0$

Here 1/2, 0, 0 can't be in GP since the common ratio is zero.

The common ratio of a GP should be non - zero.

Therefore,

$\longrightarrow\underline{\underline{n=-2}}$