The 3rd and the 9th terms of a G.p are 54 and 39366 respectively find (a) common ratio, (b) 1st term, (c) 6th term and (d) sum of the first 10 terms
In GP n-th term is:
an = a • r ⁿ⁻¹
where
a = first term
r = common ratio
a3 = a • r ²
a3 = 54
a • r ² = 5r
a = 54 / r ²
a9 = a • r ⁸
a9 = 39366
a • r ⁸ = 39366
Replace a by 54 / r ² in this equation.
54 / r ² • r ⁸ = 39366
54 • r ⁶ = 39366
Divide both sides by 54.
r ⁶ = 729
r = six root ( 729 )
r = 3
Put this value in equation:
a • r ² = 54
a • 3 ² = 54
a • 9 = 54
a = 54 / 9
a = 6
a6 = a • r ⁵ = 6 • 3 ⁵ = 6 • 243
a6 = 1458
In GP sum of the first n terms is:
Sn = a ( 1 - r ⁿ ) / ( 1 - r )
In this case n = 10 , a = 6 , r = 3
S10= 6 • ( 1 - 3¹⁰ ) / ( 1 - 3 ) =
6 • ( 1 - 59049 ) / - 2 =
6 • ( - 59048 ) / - 2 =
- 354288 / - 2
S10 = 177144
3rd term = ar^2 = 54
9th term = ar^8 = 39366
divide them:
r^6 = 729 = 3^6
so r = 3
if ar^2 = 54 and r = 3
9a = 54
a = 6
use your definitions to find a, (did that)
r (did that), ar^5, and sum(10)
My typo:
It's not correct:
a • r ² = 5r
It should be written:
a • r ² = 54
To solve this problem, we can use the formula for the nth term of a Geometric Progression (G.P), which is:
An = A * r^(n-1)
Where An is the nth term, A is the first term, r is the common ratio, and n is the position of the term.
(a) To find the common ratio (r), we can use the information given about the 3rd and 9th terms:
A3 = A * r^(3-1) = 54
A9 = A * r^(9-1) = 39366
Dividing the equation for A9 by the equation for A3, we get:
A9/A3 = (A * r^8) / (A * r^2) = r^6
39366/54 = r^6
Simplifying the equation gives:
729 = r^6
Taking the 6th root of both sides, we find:
r = 3
Therefore, the common ratio is 3.
(b) To find the first term (A), we can substitute the common ratio (r) into the equation for the 3rd term:
54 = A * 3^(3-1)
54 = A * 9
Dividing both sides by 9, we find:
A = 6
Therefore, the first term is 6.
(c) To find the 6th term (A6), we can substitute the common ratio (r) and first term (A) into the equation:
A6 = A * r^(6-1) = 6 * 3^5
A6 = 6 * 243
A6 = 1458
Therefore, the 6th term is 1458.
(d) To find the sum of the first 10 terms, we can use the formula for the sum of a Geometric Progression:
Sn = A * (r^n - 1) / (r - 1)
Substituting the values, we get:
S10 = 6 * (3^10 - 1) / (3 - 1)
S10 = 6 * (59049 - 1) / 2
S10 = 6 * 59048 / 2
S10 = 177144
Therefore, the sum of the first 10 terms is 177144.