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arun8408
The series a1,a2,… is a geometric progression. If a4=80 and a5=160, what is the value of a1?
\[\large a _{n} = a _{1} r ^{n-1} \] 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.
\[\dfrac{160}{80} = r\]
nth term of a GP= ar^{n-1} use this and write the 5th and 4th terms ,you can solve them
\[r = 2\]You can now easily determine the first term by continuous division.
You can find a1 by either just dividing by r until you reach a1, or using \[\large a _{4} = 80 = a _{1} r ^{4-1} \] \[\large a _{5} = 160 = a _{1} r ^{5-1} \]
wat value do i substitute for r?
\[\large r = \frac{ a _{5} }{ a _{4} }\]
r should be 2, 160/80 = 2 \[\large a _{4} = 80 = a _{1} * 2 ^{4-1}\] \[\large 80 = a _{1}*2 ^{3}\]
ya.. i got it.. 10 is d ans.. thank u so much
What comes next in this sequence: 1,2,6,24,120,−−?
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.