In the page geometric sequence worksheet solution4 you are going to see solution of each questions from the geometric series worksheet.

(9) 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) x a x a r = 216

a³ = 6³

a = 6

sum of their product in pairs = 156

(a/r) x a + (a) x (ar) + (a r) x (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

3 r² + 3 r + 3 - 13 r = 0

3 r² - 10 r + 3 = 0

(3r -1) (r-3) = 0

3 r - 1 = 0 r - 3 = 0

r = 1/3 r = 3

a/r = 6/(1/3) a/r = 6/3

= 6x(3/1) = 2

= 18

a = 6 a = 6

ar = 6(1/3) ar = 6(3)

= 2 = 18

Therefore the three terms are 18,6,2 or 2,6,18

(10) 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

(7 x 4)/7 = a²

a² = 4

a = ± 2

a = 2,-2

Substitute a = 2 in the in the first equation

[(1/r) + 1 + r] = 7/2

(1 + r + r²)/r = 7/2

2r² + 2 r + 2 = 7 r

2r² + 2 r - 7 r + 2 = 0

2r² - 5 r + 2 = 0

(2r - 1) (r - 2) = 0

2r - 1 = 0 r - 2 = 0

2 r = 1 r = 2

r = 1/2

a = 2 r = 1/2

Therefore the three terms are 4, 2 , 1 or 1 , 2 , 4

- Geometric sequence worksheet
- Geometric series worksheet
- Special series
- Sequence
- Arithmetic progression
- Arithmetic series
- Geometric progression
- Geometric series

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