source: ncert_exercises
topic: algebra
class_level: class_10
chapter: ch5
exercise: ex5.4
difficulty: advanced
Class 10 | Ch5: Arithmetic Progressions | Exercise 5.4
Q1. Which term of AP 121,117,113,... is first negative term? Answer: a=121, d=-4. aₙ<0 → 121+(n-1)(-4)<0 → 125-4n<0 → n>31.25. First negative term = 32nd term. ✅
Q2. Sum of 3rd and 7th terms of AP is 6 and their product is 8. Find sum of first 16 terms. Answer: a₃+a₇=(a+2d)+(a+6d)=2a+8d=6 → a+4d=3. a₃×a₇=8. Let a+4d=3 → mid term=3. (3-2d)(3+2d)=8 → 9-4d²=8 → d²=1/4 → d=±1/2. If d=1/2: a=1. S₁₆=16/2(2+15×0.5)=8×9.5=76. If d=-1/2: a=5. S₁₆=8(10-7.5)=20. Both valid. ✅
Q3. A ladder has rungs 25cm apart, 250cm tall. Bottom rung=45cm, top rung=25cm. Length of rungs forms AP. Find total length of wood required for rungs. Answer: n=(250/25)+1=11 rungs. a=45, l=25. S=11/2(45+25)=11×35=385cm. ✅
Q4. Houses in a row numbered 1 to 49. Find value of x if sum of house numbers left of house x = sum right. Answer: Sum of 1 to (x-1) = Sum of (x+1) to 49. x(x-1)/2 = 49×50/2-x(x+1)/2. x²-x=1225-x²-x. 2x²=1225 → x=√612.5. Not integer — try x=35: Left sum=35×34/2=595, Right=49×50/2-35×36/2=1225-630=595. x=35. ✅
Q5. Small terrace at a football ground comprises 15 steps, each of width 50m. First step 0.25m high, each subsequent step 0.25m higher. Find total volume of concrete required. Answer: Heights: 0.25, 0.5, 0.75,...,3.75m (AP with a=0.25,d=0.25,n=15). Each step volume=50×0.25×width. But: Step k has height k×0.25 and width (50m). Volume of step k=50×0.25×0.25×k... Standard: Total volume=sum of (length×width×height of each step)=50×0.25×S₁₅ where S₁₅=15/2(0.25+3.75)=30. Total=50×0.25×30=375 m³. ✅