Question 13 - End-of-Chapter Exercises - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress

Transcript

Question 13 The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP. Let the three terms in G.P. be 𝒂/𝒓, a, ar Given that Sum of first three terms of G.P. = 26 𝒂/𝒓 + a + ar = 26 Also, Sum of their squares = 364 (𝒂/𝒓)^𝟐+𝒂^𝟐+(𝒂𝒓)^𝟐=𝟑𝟔𝟒 Since sum and sum of squares is given We use the formula (𝐱+𝐲+𝐳)^𝟐=𝐱^𝟐+𝒚^𝟐+𝒛^𝟐+𝟐𝒙𝒚+𝟐𝒙𝒛+𝟐𝒛𝒙 Now, (𝒂/𝒓+𝒂+𝒂𝒓)^𝟐 =(𝒂/𝒓)^𝟐+𝒂^𝟐+(𝒂𝒓)^𝟐+𝟐 ×𝒂/𝒓 × 𝒂+𝟐 ×𝒂/𝒓 × 𝒂𝒓+𝟐 × 𝒂𝒓 × 𝒂 Putting values form (1) and (2) 𝟐𝟔^𝟐=𝟑𝟔𝟒+𝟐 ×𝒂^𝟐/𝒓+𝟐𝒂^𝟐+𝟐𝒂^𝟐 𝒓 676=364+2 × 𝑎(𝒂/𝒓+𝒂+𝒂𝒓) Putting (𝑎/𝑟+𝑎+𝑎𝑟)=26 from (1) 𝟔𝟕𝟔=𝟑𝟔𝟒+𝟐 × 𝒂 × 𝟐𝟔 676−364=52 × 𝑎 312=52 × 𝑎 312/52=𝑎 𝒂=𝟑𝟏𝟐/𝟓𝟐 𝑎=156/26 𝑎=78/13 𝒂=𝟔 Putting a = 6 in (1) 𝑎/𝑟+𝑎+𝑎𝑟=26 6/𝑟+6+6𝑟=26 6/𝑟+6𝑟=26−6 6/𝑟+6𝑟=20 (6 + 6𝑟 × 𝑟)/𝑟=20 6+6𝑟^2=20𝑟 𝟔𝒓^𝟐−𝟐𝟎𝒓+𝟔=𝟎 Factorising by splitting the middle term 6𝑟^2−𝟏𝟖𝒓−𝟐𝒓+6=0 6𝑟(𝑟−3)−2(𝑟−3)=0 (𝟔𝒓−𝟐)(𝒓−𝟑)=𝟎 Splitting the middle term method We need to find two numbers whose Sum = –20 Product = 6 × 6 = 36 Since sum is negative but product is positive. It means both numbers are negative Solving this Now we have a = 6 r = 𝟏/𝟑 & r = 𝟑 We need to find the terms 6r – 2 = 0 6r = 2 r = 2/6 r = 𝟏/𝟑 r – 3 = 0 r = 3 For a = 6 and r = 𝟏/𝟑 Since our terms are𝒂/𝒓, a, ar a/r = 6/(1/3) = 6 × 3 = 18 a = 6 ar = 6 × 1/3 = 2 Hence, the three term of G.P. are 18, 6, 2 For a = 6 and r = 3 Since our terms are𝒂/𝒓, a, ar a/r = 6/3 = 2 a = 6 ar = 6 × 3 = 18 Hence the three term of G.P. are 2, 6, 18 Hence first three terms of G.P. are 18, 6, 2 or 2, 6, 18