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1st term = 1 2nd term = x 3rd term = 1+x 4th term = 2 x + 2 ........... 7th term = 16x+16 Each term after the third term is the sum of all the previous terms 16x+16=1000 so x = 61.5

We are given the relation:

$\boxed{a_n=\displaystyle\sum_{i=1}^{n-1} a_i} \rightarrow (A)$ .

By splitting our sum:

$a_n=\displaystyle\sum_{i=1}^{n-2} a_i+a_{n-1}$ .

By applying $(A)$ to $a_{n-1}$ :

$a_n=\displaystyle\sum_{i=1}^{n-2} a_i+\displaystyle\sum_{i=1}^{n-2} a_i=2\displaystyle\sum_{i=1}^{n-2} a_i$ .

Repeating these steps several times:

$a_n=2\displaystyle\sum_{i=1}^{n-2} a_i=2(\displaystyle\sum_{i=1}^{n-3} a_i+a_{n-2})$

$=2(\displaystyle\sum_{i=1}^{n-3} a_i +\displaystyle\sum_{i=1}^{n-3} a_i)=2^2\displaystyle\sum_{i=1}^{n-3} a_i$

...

Obviously, we conclude:

$\boxed{a_n= 2^{k-1}\displaystyle\sum_{i=1}^{n-k} a_i } \rightarrow (B)$

Now, since we are given the seventh and the first terms of the sequence, and we are asked about the second term, substitute $n=7$ and $k=5$ in $(B)$ :

$a_7= 16\displaystyle\sum_{i=1}^{2} a_i= 16a_1+16a_2 <=> \boxed{a_2=61.5}$

For the sake of completeness, $(B)$ can be proved by induction:

For $k=1$ ,

$a_n= \displaystyle\sum_{i=1}^{n-1} a_i$ which is $(A)$ , Hence true.

replace $k$ by $k+1$ ,

$a_n= 2^{k}\displaystyle\sum_{i=1}^{n-k-1} a_i= 2^{k-1}2\displaystyle\sum_{i=1}^{n-k-1} a_i$

$= 2^{k-1}(\displaystyle\sum_{i=1}^{n-k-1} a_i+\displaystyle\sum_{i=1}^{n-k-1} a_i)$

Applying $(A)$ to $\displaystyle\sum_{i=1}^{n-k-1} a_i$ :

$a_n=2^{k-1}(\displaystyle\sum_{i=1}^{n-k-1} a_i+a_{n-k})=2^{k-1}\displaystyle\sum_{i=1}^{n-k} a_i$ and hence proved.

let the second term be x. Thus,the third term is 1+x . fourth term is 1+x+(1+x)=2+2x and so on seventh term is 16+16x . Thus 16+16x=1000 which gives x=61.5