Arithmetic Progressions

40

Step 1: Identify a=2, d=3, n=5. Step 2: Use sum formula Sn = n/2[2a + (n-1)d]. Step 3: Substitute: S5 = 5/2[2×2 + (5-1)×3]. Step 4: Calculate: S5 = 5/2[4 + 12] = 5/2 × 16 = 5 × 8 = 40. Step 5: Verify by adding terms: 2+5+8+11+14 = 40.

7

Step 1: Given t3=11 and t7=19, find first term a. Step 2: Write equations: t3 = a + 2d = 11 and t7 = a + 6d = 19. Step 3: Subtract first from second: (a + 6d) - (a + 2d) = 19 - 11, so 4d = 8, giving d = 2. Step 4: Substitute d=2 into first equation: a + 2×2 = 11, so a + 4 = 11. Step 5: Therefore a = 7.

270

Step 1: Given a=10, l=50, n=9, find sum. Step 2: Use formula Sn = n/2(a + l) when last term is known. Step 3: Substitute: S9 = 9/2(10 + 50). Step 4: Calculate: S9 = 9/2 × 60 = 9 × 30 = 270. Step 5: This formula is easier when the last term is given directly.

8

Step 1: Given a=3, d=4, Sn=120, find n. Step 2: Use formula 120 = n/2[2×3 + (n-1)×4]. Step 3: Simplify: 120 = n/2[6 + 4n - 4] = n/2[2 + 4n] = n(1 + 2n). Step 4: Solve: n + 2n² = 120, so 2n² + n - 120 = 0. Step 5: Factor: (2n + 15)(n - 8) = 0, giving n = 8 (positive solution).