JEE Main 2012 — Mathematics Algebra

Let $a, a r, a r^2, a r^3, a r^4, a r^5$a,ar,ar2,ar3,ar4,ar5 be six terms of a G.P. where ' $a$a ' is first term and $\mathrm{r}$r is common ratio. According to given conditions, we have $a r^3-a=5 \Rightarrow a\left(r^3-1\right)=52$ar3−a=5⇒a(r3−1)=52 and $a+a r+a r^2=26$a+ar+ar2=26 $\Rightarrow a\left(1+r+r^2\right)=26$⇒a(1+r+r2)=26 To find: $a\left(1+r+r^2+r^3+r^4+r^5\right)$a(1+r+r2+r3+r4+r5) Consider $$\begin{aligned} & a\left[1+r+r^2+r^3+r^4+r^5\right] \\ & =a\left[1+r+r^2+r^3\left(1+r+r^2\right)\right] \\ & =a\left[1+r+r^2\right]\left[1+r^3\right] \end{aligned}$$​a[1+r+r2+r3+r4+r5]=a[1+r+r2+r3(1+r+r2)]=a[1+r+r2][1+r3]​ Divide (1) by (2), we get $$\frac{r^3-1}{1+r+r^2}=2$$1+r+r2r3−1​=2 we know $r^3-1=(r-1)\left(1+r+r^2\right)$r3−1=(r−1)(1+r+r2) $$\begin{aligned} & \therefore r-1=2 \Rightarrow r=3 \text { and } a=2 \\ & \therefore \quad a\left(1+r+r^2+r^3+r^4+r^5\right) \\ & \quad=a\left(1+r+r^2\right)\left(1+r^3\right) \\ & \quad=2(1+3+9)(1+27) \\ & \quad=26 \times 28=728 \end{aligned}$$​∴r−1=2⇒r=3 and a=2∴a(1+r+r2+r3+r4+r5)=a(1+r+r2)(1+r3)=2(1+3+9)(1+27)=26×28=728​