Question

Find the sum of the corresponding to the planets -

• A. $313$
• B. $396$
• C. $380$
• D. $413$

Answer 1

The correct option is D $413$

Mercury$\left(P_1\right)=0+4=4$

Venus$\left(P_2\right)=3+4=7$

Earth$\left(P_3\right)=6+4=10$

Mars$\left(P_4\right)=12+4=16$

Saturn$\left(P_7\right)=96+4$

As we can see $4$ is added to each planet distance and from Venus the next digit is multiplied by $2$.

$=4\times n+0+3+6+12+.....96$

Total there are $8$ planets.

$=4\times 8+0+3\times 2^0+3\times 2^1+3\times 2^2......3\times 2^5+3\times 2^6$

Sum of $GP$ when series$=a+ar+a{r}^2+.......a{r}^{n-1}$

$=\frac{a(r^n-1)}{r-1}$ where $r>1$;

So sum,when $a=3,r=2$ and $n=7$

$=\frac{3(2^7-1)}{2-1}=3(128-1)=3\left(127\right)=381$

Sum$=32+381=413$

Answer $\left(D\right)$