Data:
a5 = - 48
Being r = q = 2
a1 = ?
a8 = ?
n (Number of terms)
Solving:
Formula (geometric progression)
\(a_{n} = a_{1} *q^{n-1}\)
We have:
find the 1th (a1)
\(a_{n} = a_{1} *q^{n-1}\)
\(a_{5} = a_{1} *2^{5-1}\)
\(-48 = a_{1} *2^{4}\)
\(-48 = a_{1} *16\)
\(16a_{1} = -48\)
\(a_{1} = \frac{-48}{16}\)
\(\boxed{a_{1} =-3}\)
find the 8th (a8)
\(a_{n} = a_{1} *q^{n-1}\)
\(a_{8} = -3 *2^{8-1}\)
\(a_{8}= -3 *2^{7}\)
\(a_{8}= -3 *128\)
\(\boxed{a_{8} =-384}\)
Answer:
\(\boxed{a_{1} = -3}\)
\(\boxed{a_{8} = -384}\)
