Find the Terms From Sum and product of 3 Consecutive Terms of GP :
In this section, we will learn how to find the terms from sum and product of 3 consecutive terms of a geometric progression.
We can consider the first three terms of a geometric progression as
a/r, a, ar
Let us see some examples based on the concept.
Example 1 :
If the product of three consecutive terms in G.P is 216 the sum of their product in pairs is 156, find them.
Solution :
Let the first three terms are a/r, a, ar
Product of three terms = 216
(a/r) ⋅ a ⋅ a r = 216
a3 = 63
a = 6
Sum of their product in pairs = 156
(a/r) ⋅ a + a ⋅ ar + ar ⋅ (a/r) = 156
a2 / r + a2 r + a2 = 156
a2 [ (1/r) + r + 1 ] = 156
a² [ (1+r²+r)/r] = 156
a² (r²+r+1)/r = 156
(6²/r)(r²+r+1) = 156
(r²+r+1)/r = 156/36
(r²+r+1)/r = 13/3
3(r²+r+1) = 13 r
3r² + 3r + 3 - 13r = 0
3r² - 10r + 3 = 0
(3r - 1)(r - 3) = 0
3r - 1 = 0 3r = 1 r = 1/3 |
r - 3 = 0 r = 3 |
If a = 6, then r = 1/3
a/r = 6/(1/3) ==> 18
a = 6
ar = 6(1/3) ==> 2
If a = 6, then r = 1/3
a/r = 6/3 ==> 2
a = 6
ar = 6(3) ==> 18
Hence the required three terms are 18, 6 and 2 or 2, 6, 18.
Example 2 :
Find the first three consecutive terms in G.P whose sum is 7 and the sum of their reciprocals is 7/4.
Solution :
Let the first three terms are a/r, a, ar
Sum of three terms = 7
a/r + a + a r = 7
a [(1/r) + 1 + r] = 7
[(1/r) + 1 + r] = 7/a -----------(1)
Sum of their reciprocals = 7/4
(r/a) + (1/a) + (1/ar) = 7/4
(1/a)[r + 1 + (1/r)] = 7/4
(1/a)(7/a) = 7/4
7/a² = 7/4
a2 = 4
a = 2
By applying a = 2 in the in (1), we get
[(1/r) + 1 + r] = 7/2
(1 + r + r2)/r = 7/2
2r2 + 2r + 2 = 7r
2r² + 2r - 7r + 2 = 0
2r² - 5r + 2 = 0
(2r - 1) (r - 2) = 0
2r - 1 = 0 r = 1/2 |
r - 2 = 0 r = 2 |
If a = 2 and r = 1/2
a/r = 2/(1/2) ==> 4
a = 2
ar = 2(1/2) ==> 1
Therefore the three terms are 4, 2 , 1 or 1 , 2 , 4
After having gone through the stuff given above, we hope that the students would have understood how to find the terms from the sum and product of 3 terms of geometric progression.
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