Ch 5: AP - Sum of n Terms
Master the Sum formulas [ Sn = n/2 (2a + (n-1)d) ] and tackle the most complex board word problems.
Exercise 5.3 Solutions
š” Note: If the last term (l) is given, use the shortcut formula: Sn = n/2 (a + l)
Q1. Find the sum of the following APs: (i) 2, 7, 12, ... to 10 terms.
Solution:
Here, a = 2, d = 7 - 2 = 5, n = 10.
Formula: Sn = (n/2) [2a + (n-1)d]
Sāā = (10/2) [2(2) + (10 - 1)5]
Sāā = 5 [4 + 9(5)]
Sāā = 5 [4 + 45]
Sāā = 5 Ć 49 = 245.
Q3. In an AP: (i) given a = 5, d = 3, an = 50, find n and Sn.
Solution:
Step 1: Find n using an formula.
an = a + (n-1)d
50 = 5 + (n-1)3
45 = 3(n-1) ā n-1 = 15 ā n = 16.
Step 2: Find Sn using the shortcut formula (since an is the last term l).
Sn = (n/2) (a + l)
Sāā = (16/2) (5 + 50)
Sāā = 8 Ć 55 = 440.
Q12. Find the sum of the first 40 positive integers divisible by 6.
Solution:
The positive integers divisible by 6 are: 6, 12, 18, 24...
This forms an AP where a = 6, d = 6, and n = 40.
Sāā = (40/2) [2(6) + (40 - 1)6]
Sāā = 20 [12 + 39(6)]
Sāā = 20 [12 + 234]
Sāā = 20 Ć 246 = 4920.
Advanced Word Problems (Ex 5.3 & Ex 5.4 Optional)
Q16. A sum of ā¹700 is to be used to give seven cash prizes to students of a school. If each prize is ā¹20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the 1st prize be a. Since each prize is ā¹20 less, the common difference d = -20.
Total sum (Sā) = 700, and n = 7.
Formula: Sā = (n/2) [2a + (n-1)d]
700 = (7/2) [2a + (7-1)(-20)]
700 Ć 2 / 7 = 2a + 6(-20)
200 = 2a - 120
2a = 320 ā a = 160
The prizes are: ā¹160, ā¹140, ā¹120, ā¹100, ā¹80, ā¹60, ā¹40.
Q1. Which term of the AP: 121, 117, 113, ... is its first negative term?
Solution:
Here, a = 121, d = 117 - 121 = -4.
For the first negative term, an < 0.
a + (n-1)d < 0
121 + (n-1)(-4) < 0
121 - 4n + 4 < 0
125 < 4n
n > 125 / 4
n > 31.25
Since 'n' must be an integer, the next integer greater than 31.25 is 32.
Therefore, the 32nd term is the first negative term.