Sequences and Series

6

Step 1: This is a GP with a = 3, r = 3. Sum formula: Sn = a(rⁿ - 1)/(r - 1) = 3(3ⁿ - 1)/2. Step 2: We need Sn > 1000, so 3(3ⁿ - 1)/2 > 1000, giving 3ⁿ - 1 > 666.67, so 3ⁿ > 667.67. Step 3: Check n=5: 3⁵ = 243. S5 = 3(243-1)/2 = 3×242/2 = 363. Not exceeding 1000. Step 4: Check n=6: 3⁶ = 729. S6 = 3(729-1)/2 = 3×728/2 = 1092 > 1000. ✓ Step 5: So 6 terms are needed. Students often make arithmetic errors with powers of 3; note 3⁶ = 729, not 648.

2/3

Step 1: Let a be the first term and r be the common ratio. Then t4 = ar³ = 2/3 and t7 = ar⁶ = 16/81. Step 2: Divide t7 by t4: (ar⁶)/(ar³) = (16/81)/(2/3). Step 3: r³ = (16/81) × (3/2) = 48/162 = 8/27. Step 4: Therefore r = ∛(8/27) = 2/3. Step 5: Verify: t4 = ar³ = 2/3 and t7 = ar⁶ = ar³ × r³ = (2/3)(8/27) = 16/81. ✓ A common error is to compute t7/t4 = r⁴ instead of r³.

3

Step 1: Let first term be a and common ratio r. Sum S = a/(1-r) = 4, so a = 4(1-r). Step 2: The squares of the GP terms a², a²r², a²r⁴,... form a GP with first term a² and common ratio r². Sum of squares = a²/(1-r²) = 16/3. Step 3: a²/(1-r²) = a²/[(1-r)(1+r)] = [a/(1-r)] × [a/(1+r)] = 4 × [a/(1+r)] = 16/3. Step 4: So a/(1+r) = 4/3. Also a/(1-r) = 4. Dividing: (1-r)/(1+r) = 1/3, giving 3-3r = 1+r, so 4r = 2, r = 1/2. Step 5: Therefore a = 4(1 - 1/2) = 4 × 1/2 = 2. Wait - let me recheck: a/(1-r) = 4 with r=1/2 gives a = 4×1/2 = 2. But a/(1+r) = 2/(3/2) = 4/3. ✓. Hmm, first term = 2. Let me re-ex

11

Step 1: Let the three numbers in AP be (a-d), a, (a+d). Their sum = 3a = 24, so a = 8. Step 2: Product = (a-d)(a)(a+d) = a(a² - d²) = 440. Step 3: 8(64 - d²) = 440, so 64 - d² = 55, giving d² = 9, d = ±3. Step 4: When d = 3: numbers are 5, 8, 11. When d = -3: numbers are 11, 8, 5. Step 5: In both cases, the three numbers are 5, 8, 11. The largest number is 11. A common mistake is to forget to use a as the middle term or to make errors expanding (a-d)(a+d) = a²-d².