Question

Let $a_n=n{.2}^n+n^2+n+1,bn=a_{n+1}-a_n,C_n=b_{n+1}-b_n,d_n=C_{n+1}-C_n\dots$ and
So on $Z_n=y_{n+1}-y_n\forall nϵN$

$y_2+z_2,y_3+z_3,y_4+z_4$ are in

• A. A.P
• B. G.P
• C. H.P
• D. A.G.P

Answer 1

The correct option is D A.G.P

$b_n=\left(\right(n+1)2^{n+1}+(n+1{)}^2+(n+1)+1)-(n{2}^n+n^2+n+1)$

$b_n=(n+2)2^n+2n+2$

Similarly $c_n=(n+4)2^n+2d_n=(n+6)2^n{e}_n=(n+8)2^n$

According to pattern -$x_n=(n+46)2^n{y}_n=(n+48)2^n{z}_n=(n+50)2^n$

$Sin\theta .y_n+z_n=y_{n+1}\Rightarrow$ $y_2+z_2=y_1{y}_3+z_3=y_2{y}_4+z_4=y_3$

So $y_1,y_2,y_3$ are in A.G.P