Let $S_n$ denote the sum of the cubes of the first $n$ natural numbers and $s_n$ denote the sum of the first $n$ natural numbers, then $\sum_\limits{r=1}^n \frac{S_r}{S_4}$ equals to
If $t_n$ denotes the $n$th term of the series $2+3+6+11+18+\ldots$, then $t_{50}$ is
The lengths of three unequal edges of a rectangular solid block are in GP. If the volume of the block is $216 \mathrm{~cm}^3$ and the total surface area is $252 \mathrm{~cm}^2$, then the length of the longest edge is
If $a, b$ and $c$ are in GP, then the value of $\frac{a-b}{b-c}$ is equal to ............ .
Given that, $a, b$ and $c$ are in GP.
Then, $$\frac{b}{a}=\frac{c}{b}=r$$ [constant]
$$\begin{aligned} \Rightarrow \quad & b =a r \Rightarrow c=b r \\ \therefore \quad & \frac{a-b}{b-c} =\frac{a-a r}{a r-b r}=\frac{a(1-r)}{r(a-b)}=\frac{a(1-r)}{r(a-a r)} \\ & =\frac{a(1-r)}{a r(1-r)}=\frac{1}{r} \\ \therefore \quad & \frac{a-b}{b-c} =\frac{1}{r}=\frac{a}{b} \text { or } \frac{b}{c} \end{aligned}$$
The sum of terms equidistant from the beginning and end in an AP is equal to ........... .
Let AP be $a, a+d, a+2 d \cdots a+(n-1) d$
$$\begin{array}{l} \therefore \quad a_1+a_n & =a+a+(n-1) d \\ & =2 a+(n-1) d \quad \text{.... (i)}\\ \text { Now, } \quad a_2+a_{n-1} & =(a+d)+[a+(n-2) d] \\ & =2 a+(n-1) d \\ & a_2+a_{n-1} =a_1+a_n \quad \text{[using Eq. (i)}\\ & a_3+a_{n-2} =(a+2 d)+[a+(n-3) d] \\ & =2 a+(n-1) d \\ & =a_1+a_n \quad \text{[using Eq. (i)} \end{array}$$
Follow this pattern, we see that the sum of terms equidistant from the beginning and end in an AP is equal to [first term + last term].