If each term of a geometric progression a_1,a_2,a_3,… with a_1= 1 8 and a_2≠ a_1, is the arithmetic mean of the next two terms and S_n=a_1+a_2+…

Let $\mathrm{GP}:a,\mathrm{ar},{\mathrm{ar}}^{2},{\mathrm{ar}}^{3},...$GP:a,ar,ar2,ar3,...

Now, given: $2\mathrm{ar}={\mathrm{ar}}^{2}+{\mathrm{ar}}^{3}$2ar=ar2+ar3

$$\Rightarrow 2=r+{r}^{2}$$⇒2=r+r2

$$\Rightarrow {r}^{2}+r-2=0$$⇒r2+r−2=0

$$\Rightarrow (r+2)(r-1)=0$$⇒(r+2)(r−1)=0

$$\Rightarrow r=-2,1[\text{Since}\mathrm{GP}\text{is not constant so }r=1\text{ is rejected}]$$⇒r=−2,1[SinceGPis not constant so r=1 is rejected]

$$\Rightarrow r=-2$$⇒r=−2

$$\Rightarrow {S}_{20}-{S}_{18}=\frac{a(1-{r}^{20})}{1-r}-\frac{a(1-{r}^{18})}{1-r}$$⇒S20​−S18​=1−ra(1−r20)​−1−ra(1−r18)​

$$\Rightarrow {S}_{20}-{S}_{18}=\frac{a[{r}^{18}-{r}^{20}]}{1-r}$$⇒S20​−S18​=1−ra[r18−r20]​

$$\Rightarrow {S}_{20}-{S}_{18}=\frac{[{2}^{18}-{2}^{20}]}{8\times 3}$$⇒S20​−S18​=8×3[218−220]​

$$\Rightarrow {S}_{20}-{S}_{18}=\frac{{2}^{18}(1-4)}{24}$$⇒S20​−S18​=24218(1−4)​

$$\Rightarrow {S}_{20}-{S}_{18}=-{2}^{15}$$⇒S20​−S18​=−215