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].
The third term of a GP is 4 , the product of the first five terms is ............. .
It is given that, $T_3=4$
Let $a$ and $r$ the first term and common ration, respectively.
$$\begin{aligned} \text{Then, }\quad a r^2 & =4 \quad \text{.... (i)}\\ & \text { Product of first } 5 \text { terms }=a r \cdot a r \cdot a r^2 \cdot a r^3 \cdot a r^4\\ & =a^5 r^{10}=\left(a r^2\right)^5=(4)^5\quad \text{[using Eq. (i)]} \end{aligned}$$
Two sequences cannot be in both AP and GP together.
Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.
Any term of an AP (except first) is equal to half the sum of terms which are equidistant from it.