NCERT MATH CLASS 10 | NCERT MATH CLASS 10 EX 5.3 SOLUTIONS

NCERT MATH CLASS 10 | NCERT MATH CLASS 10 EX 5.3 SOLUTIONS Quick formula recap (must-know) nth term of an AP: 𝑎 𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 a n ​ =a+(n−1)d where 𝑎 a = first term, 𝑑 d = common difference. Sum of first n terms: 𝑆 𝑛 = 𝑛 2 ( 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ) S n ​ = 2 n ​ (2a+(n−1)d) or 𝑆 𝑛 = 𝑛 2 ( 𝑎 + 𝑎 𝑛 ) S n ​ = 2 n ​ (a+a n ​ ). Typical Question 1 — Find the n-th term Problem: Find the 20th term of the AP 3, 7, 11, ... Solution: First term 𝑎 = 3 a=3. Common difference 𝑑 = 7 − 3 = 4 d=7−3=4. Use 𝑎 𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 a n ​ =a+(n−1)d. For 𝑛 = 20 n=20: 𝑎 20 = 3 + ( 20 − 1 ) ⋅ 4 = 3 + 19 ⋅ 4 a 20 ​ =3+(20−1)⋅4=3+19⋅4 Compute 19 × 4 = 76 19×4=76. So 𝑎 20 = 3 + 76 = 79. a 20 ​ =3+76=79. Answer: 79. 79. Typical Question 2 — Given two terms, find a and d Problem: In an AP, 𝑎 5 = 12 a 5 ​ =12 and 𝑎 10 = 27 a 10 ​ =27. Find 𝑎 a and 𝑑 d. Solution: We know 𝑎 𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 a n ​ =a+(n−1)d. From 𝑎 5 = 𝑎 + 4 𝑑 = 12 a 5 ​ =a+4d=12. … (1) From 𝑎 10 = 𝑎 + 9 𝑑 = 27 a 10 ​ =a+9d=27. … (2) Subtract (1) from (2): ( 𝑎 + 9 𝑑 ) − ( 𝑎 + 4 𝑑 ) = 27 − 12 (a+9d)−(a+4d)=27−12 So 5 𝑑 = 15 5d=15 → 𝑑 = 3. d=3. Now substitute back: 𝑎 + 4 ⋅ 3 = 12 a+4⋅3=12 → 𝑎 + 12 = 12 a+12=12 → 𝑎 = 0. a=0. Answer: 𝑎 = 0 , 𝑑 = 3. a=0, d=3. Typical Question 3 — Find number of terms Problem: Find n if 𝑎 𝑛 = 98 a n ​ =98 for the AP 2, 5, 8, ... Solution: Here 𝑎 = 2 , 𝑑 = 5 − 2 = 3. a=2, d=5−2=3. Use 𝑎 𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 a n ​ =a+(n−1)d. Put 𝑎 𝑛 = 98 a n ​ =98: 98 = 2 + ( 𝑛 − 1 ) ⋅ 3 98=2+(n−1)⋅3 Subtract 2: 96 = 3 ( 𝑛 − 1 ) 96=3(n−1). Divide by 3: 32 = 𝑛 − 1 32=n−1. So 𝑛 = 33. n=33. Answer: 𝑛 = 33. n=33. Typical Question 4 — Sum of n terms Problem: Find sum of first 15 terms of the AP 4, 9, 14, ... Solution: Here 𝑎 = 4 , 𝑑 = 5. a=4, d=5. Use 𝑆 𝑛 = 𝑛 2 ( 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ) S n ​ = 2 n ​ (2a+(n−1)d) with 𝑛 = 15 n=15: 𝑆 15 = 15 2 ( 2 ⋅ 4 + ( 15 − 1 ) ⋅ 5 ) = 15 2 ( 8 + 14 ⋅ 5 ) S 15 ​ = 2 15 ​ (2⋅4+(15−1)⋅5)= 2 15 ​ (8+14⋅5) Compute 14 ⋅ 5 = 70 14⋅5=70. So inside bracket 8 + 70 = 78. 8+70=78. Then 𝑆 15 = 15 2 ⋅ 78. S 15 ​ = 2 15 ​ ⋅78. First compute 15 × 78 15×78. Calculate: 78 × 15 = 78 × ( 10 + 5 ) = 780 + 390 = 1170. 78×15=78×(10+5)=780+390=1170. So 𝑆 15 = 1170 2 = 585. S 15 ​ = 2 1170 ​ =585. Answer: 585. 585. Typical Question 5 — Mixed type (find d from sums) Problem: Sum of first 7 terms of an AP is 49 and sum of first 14 terms is 196. Find 𝑎 a and 𝑑 d. Solution idea: Use formula for 𝑆 𝑛 S n ​ and set up two equations. (You can show full algebraic steps in your answer sheet.)