# CLASS-10ARITHMETIC 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.)