Question

Can anyone help me? There is an geometrical progression. The sum of first members of the sequance can be calculated according to the formula: \[S _{n}=\frac{ 1 }{ 4 }(2^{n}-1)\]

Need to find a) sixth member of a sequance (\b_{6}\) and "q"

So, i think I can do by this:

\[b _{6}=S _{6}-S _{5}=(\frac{ 1 }{ 4 }(2^{6}-1))-((\frac{ 1 }{ 4 }(2^{5}-1))=15,75-8,25=7,5\]
but the answwer should be 7, any suggestions?

Answer 1

@saifoo.khan

Answer 2

Maybe you don't understand my labelling? I know it is quite different in other countries?

Answer 3

@hartnn any ideas?:/

Answer 4

S6-S5 = 1/4 {2^6-1-2^5+1} = 1/4 (32) = 8 .....

Answer 5

Oh, right

((14(2^5−1))=7.75 and the answer should be 8... Well, maybe mistake in the answer sheet:/

Answer 6

even i think so...

Answer 7

thank you!