Question 4 6:46:29 Four numbers are given. The first three numbers are terms of a geometric sequence and the last three numbers are terms of an arithmetic sequence. The sum of the first and the last number is 21 and the sum of the middle two numbers is 18. Find those numbers?
Answers
Solution :-
Let four numbers are a, b, c and d respectively .
so, a, b and c are in GP,
→ b² = ac ----- Eqn.(1)
and, b, c and d are in AP,
→ 2c = (b + d) ----- Eqn.(2)
now, given that,
→ a + d = 21 ------ Eqn.(3)
→ b + c = 18 ------- Eqn.(4)
adding Eqn.(3) and Eqn.(4) ,
→ (a + b + c + d) = 21 + 18 = 39
putting value of Eqn.(2)
→ a + c + 2c = 39
→ a + 3c = 39
→ a = (39 - 3c) ------ Eqn.(5)
putting value of a in Eqn.(1) ,
→ b² = c(39 - 3c)
putting value of c from Eqn.(4) ,
→ (18 - c)² = c(39 - 3c)
→ 324 + c² - 36c = 39c - 3c²
→ 4c² - 75c + 324 = 0
→ 4c² - 48c - 27c + 324 = 0
→ 4c(c - 12) - 27(c - 12) = 0
→ (c - 12)(4c - 27) = 0
→ c = 12 or (27/4)
taking c = 12 :-
→ b + c = 18 (Eqn.4)
→ b + 12 = 18
→ b = 18 - 12 = 6
and,
→ a = 39 - 3c (Eqn.5)
→ a = 39 - 3*12 = 39 - 36 = 3
and,
→ a + d = 21 (Eqn.3)
→ 3 + d = 21
→ d = 21 - 3 = 18 .
therefore, four numbers are 3, 6, 12 and 18 .
taking c = (27/4)
→ b = 18 - (27/4) = (45/4)
→ a = 39 - (81/4) = (75/4)
→ d = 21 - (75/4) = (9/4)
therefore, four numbers are (75/4), (45/4), (27/4) and (9/4) .
Learn more :-
evaluate the expression given by 83 - 81 + 87 - 85 +__________ + 395 - 393 + 399 - 397
https://brainly.in/question/14081691
If the nth term of an AP is (2n+5),the sum of first10 terms is
https://brainly.in/question/23676839