Question

The series a1,a2,… is a geometric progression. If a4=80 and a5=160, what is the value of a1?

Answer 1

\[\large a _{n} = a _{1} r ^{n-1} \] is the formula for a geometric progression, r being the common ratio, and a1 being the first term, an being the n-th term.

r is the common ratio, so if you divide a5 by a4, you'll find r.

Then you can use \[\large a _{4} = 80 = a _{1} r ^{4-1}\] to find a1.

Answer 2

\[\dfrac{160}{80} = r\]

Answer 3

nth term of a GP= ar^{n-1}

use this and write the 5th and 4th terms ,you can solve them

Answer 4

\[r = 2\]You can now easily determine the first term by continuous division.

Answer 5

You can find a1 by either just dividing by r until you reach a1, or using

\[\large a _{4} = 80 = a _{1} r ^{4-1} \] \[\large a _{5} = 160 = a _{1} r ^{5-1} \]

Answer 6

wat value do i substitute for r?

Answer 7

\[\large r = \frac{ a _{5} }{ a _{4} }\]

Answer 8

a1 . r^3 = a1 . 16??

Answer 9

r should be 2, 160/80 = 2

\[\large a _{4} = 80 = a _{1} * 2 ^{4-1}\] \[\large 80 = a _{1}*2 ^{3}\]

Answer 10

ya.. i got it.. 10 is d ans.. thank u so much

Answer 11

What comes next in this sequence:

1,2,6,24,120,−−?

Answer 12

There's a pattern:
first term: 1x1 = 1
second term: 2x1 = 2
third term: 3x2 = 6
fourth term: 4x6 = 24
fifth term: 5x24 = 120
sixth term: ? x120 = ?

Look at the pattern of multiples on the left.