Asked by dennis
the common ratio of a g.p is 2.if the 5th term is greater than the first term by 45, find the 5tth term
Answers
Answered by
drwls
Is g.p. supposed to mean geometric progression? You should not assume that we know that.
The fifth term will be 2^3 = 8 times the second term
a5 = 8 a2
a5 = a2 + 45
0 = 7 a2 -45
a2 = 45/7
a3 = 90/7
a4 = 180/7
a5 = 360/7
Check: a5 - a2 = 315/7 = 45
The fifth term will be 2^3 = 8 times the second term
a5 = 8 a2
a5 = a2 + 45
0 = 7 a2 -45
a2 = 45/7
a3 = 90/7
a4 = 180/7
a5 = 360/7
Check: a5 - a2 = 315/7 = 45
Answered by
Adorbs
U didn't really explain it well . I needed a full and understandable answer.
Answered by
MfonAbasi
If the first term of a GP is ‘a’ and the common ratio is r, then the n’th term is: ar^(n − 1).
5th term is greater than 1st term by 45. ar^4 - a = 45.
Substitute r = 2 in the expression.
a*2^4 - a = 45
16a - a = 45
15 a = 45
a = 3
5th term = ar^(5 -1) = 3 * 2^4 = 48
5th term is greater than 1st term by 45. ar^4 - a = 45.
Substitute r = 2 in the expression.
a*2^4 - a = 45
16a - a = 45
15 a = 45
a = 3
5th term = ar^(5 -1) = 3 * 2^4 = 48
Answered by
Esther
a × r^4 - a = 45
Where a= 2
a × 2^4 - a =45
a × 16 - a =45
16a - a= 45
15a =45
Divide through by 15
a = 3
Tn = ar^n-1
Where n= 5 ;a= 3 ; r= 2
T5 = 3 × 2^5 -1
T5 = 3 × 2^4
T5 = 48
Where a= 2
a × 2^4 - a =45
a × 16 - a =45
16a - a= 45
15a =45
Divide through by 15
a = 3
Tn = ar^n-1
Where n= 5 ;a= 3 ; r= 2
T5 = 3 × 2^5 -1
T5 = 3 × 2^4
T5 = 48
Answered by
Remilekun
Very good