LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

ARITHMETIC PROGRESSION - PROBLEM & SOLUTION

**PROBLEM & SOLUTION OF AN ARITHMETIC PROGRESSION (A.P.) -**

**Example.1) Which term of the A.P. 4, 9, 14, 19,............... is 79 ?
**

**Ans.) The given A.P. is 4, 9, 14, 19,...............
**

**Its 1st term, a = 4 and common difference, d = 9 – 4 = 5.
**

**Let, its eth term be 79. Then –
**

**As per the rules, Tₑ = a + (e – 1) d
**

**When, Tₑ = 79, a = 4, and d = 5
**

**So, 79 = 4 + (e – 1) 5
**

**=> 5 (e – 1) = 75
**

**=> (e – 1) = 15
**

**=> e = 15 + 1 = 16
**

**Hence, the 16th term of the given A.P. is 79 (Ans.)**

**Example.2) Which term of the A.P. 5, 12, 19, 26, 33,………………will be 35 more than its 12th term ?
**

**Ans.) The given A.P. is 5, 12, 19, 26, 33,…………….
**

**Its 1st term, a = 5, and common difference, d = 12 – 5 = 19 – 12 = 26 – 19 = 33 – 26 = 7 (constant)
**

**First we to find out 12th term, so –
**

**As per the rules, Tₑ = a + (e – 1) d
**

**When, e = 12, a = 5, and d = 7
**

** T₁₂ = 5 + (12 – 1) 7
**

** = 5 + 77 = 82
**

**Now, the required term is = (82 + 35) = 117
**

**Let the eth term be 117, then
**

** As per the rules, Tₑ = a + (e – 1) d
**

**When, Tₑ = 117, a = 5, and d = 7
**

** 117 = 5 + (e – 1) 7
**

**=> 7 (e – 1) = 112
**

**=> (e – 1) = 16
**

**=> e = 16 + 1 = 17
**

**Hence, the 17th term is the required term. (Ans.)**

**Example.3) How many terms of the A.P. 7, 11, 15, 19, 23, ………………. must be taken to get the sum 250?
**

**Ans.) The given A.P. is 7, 11, 15, 19, 23,……………….
**

**Here, a = 7, d = 11 – 7 = 15 – 11 = 19 – 15 = 23 – 19 = 4 (constant), Sₑ = 250
**

**As per the rules,
**

** e
**

**Using the formula, Sₑ = ------ [2a + (e – 1) d], we get -
**

** 2
**

** e
**

** => 250 = ------- [(2 X 7) + (e – 1) 4]
**

** 2
**

** => 500 = e (14 + 4e – 4)
**

** => 500 = 4e² + 10e
**

** => 4e² + 10e – 500 = 0
**

** => 2e² + 5e – 250 = 0
**

** => 2e² + (25 – 20) e – 250 = 0
**

** => 2e² + 25e – 20e – 250 = 0
**

** => e (2e + 25) – 10 (2e + 25) = 0
**

** => (2e + 25) (e – 10) = 0
**

** => (2e + 25) = 0 or (e – 10) = 0
**

** => e = - 25/2, or e = 10
**

** => e = 10 [because, e ≠ - 25/2]
**

**Hence the required number of the terms is 10. (Ans.)
**

**Example.4) How many terms of the A.P. 17, 15, 13, 11, 9,………….must be added to get the sum 72. Explain the double answer.
**

**Ans.) Hence, a = 17, and d = (15 – 17) = (13 – 15) = (11 – 13) = (9 – 11) = - 2
**

**Let, the sum of e terms be 72. So, Sₑ = 72.
**

** Then –
**

** e
**

**Using the formula, Sₑ = ------- [2a + (e – 1) d], we get -
**

** 2
**

** e**

** => 72 = ------ [(17 X 2) + (e – 1) (- 2)]
**

** 2
**

** => 144 = e [34 + (-2e + 2)]
**

** => 144 = e (36 – 2e)
**

** => 72 = 18e - e²
**

** => e² - 18e + 72 = 0
**

** => e² - (12 + 6)e + 72 = 0
**

** => e² - 12e – 6e + 72 = 0
**

** => e(e – 12) – 6(e – 12) = 0
**

** => (e – 12) (e – 6) = 0
**

** => (e – 12) = 0 or (e – 6) = 0
**

** => e = 12 and e = 6
**

**So, sum of first 6 terms = sum of first 12 terms = 72
**

**This means that the sum of all terms from 7th to 12th is zero (Ans.)**

Your second block of text...