Find four numbers in gp whose sum is 85 and product is 4096
Answers
Let a/r³ , a/r , a and ar³ are the four numbers in GP where a and r are constants.
A/c to question,
sum of all numbers = 85
a/r³ + a/r + ar + ar³ = 85
a[1/r³ + 1/r + r + r³ ] = 85 -----(1)
Product of all numbers = 4096
a/r³ × a/r × a × ar³ = 4096
⇒a⁴ = (64)² = 8⁴
⇒ a = 8
Now, from equation (1),
8[1/r³ + 1/r + r + r³ ] = 85
⇒8(1 + r² + r⁴ + r^6)/r³ = 85
⇒8(1 + r² + r⁴ + r^6) = 85r³
⇒8 + 8r² + 8r⁴ + 8r^6 - 85r³ = 0
⇒8r^6 + 8r⁴ - 85r³ + 8r² + 8 = 0
⇒8r^6 - 16r^5 + 16r^5 - 32r⁴ + 40r⁴ - 80r³ - 5r³ + 10r² - 2r² + 8 = 0
⇒8r^5(r - 2) + 16r⁴(r - 2) + 40r³(r - 2) - 5r²(r - 2) - 2(r² - 4) = 0
⇒(r - 2)[8r^5 + 16r⁴ + 40r³ - 5r² - 2(r + 2)] = 0
⇒(r - 2)[8r^5 - 4r⁴ + 20r⁴ - 10r³ + 50r³ - 25r³ + 20r² - 10r + 8r - 4 ] = 0
⇒(r - 2)[4r⁴(2r - 1) + 10r³(2r - 1) + 25r²(2r - 1) + 10r(2r - 1) + 4(2r - 1)] = 0.
⇒(r - 2)(2r - 1)(4r⁴ + 10r³ + 25r² + 10r -
+ 4] = 0
we get, (r - 2) = 0 or, (2r - 1) = 0
so, r = 2 or 1/2
for r = 2, a = 8
first term = a/r³ = 8/2³ = 1
2nd term = a/r = 8/2 = 4
3rd term = ar = 8 × 2 = 16
4th term = ar³ = 8(2)³ = 64
similarly, you can take r = 1/2 , a = 8
first term = 8/(1/2)³ = 64
2nd term = 8/(1/2) = 16
3rd term = 8(1/2) = 4
4th term = 8(1/2)³ = 1
hence, numbers are 64, 16, 4 , 1
Answer:
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