Arithmetic Progression Problems with Answers for Competitive Exams

In this session explained about arithmetic progression problems like finding the n^th term , sum to first n^th terms, finding the number of terms in given sequence. . . .  etc.

Arithmetic Progression Examples with Solutions for class 10

Please go through the below link for basic concepts of Sequence and series, fundamental concepts with formulas and properties for arithmetic progression

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Arithmetic Progression real life problems

Example – 1:  Jhon put ₹ 800 into his son’s kiddy bank when he was one year old and increased the amount by 1000 every year. Find the amount of money in the kiddy bank on her on his 1st, 2nd, 3rd, 4th, . . . . .  birthday

Solution: Here a = 800 and d = 1000

General form of A.P is a, a+d, a+2d, a+3d, . . . . . . . . .

So the sequence is 800, 1800, 2800, 3800, 4800, . . . . . . .

Example – 2:  Cab/Taxi Rental Rates after each km when the fare is ₹ 30 for the first km and raise by 12 for each additional km. what will be the charge after traveling of 50km

Solution: Here a = 30 and d = 12 and nth term is 50

nth term = a + ( n-1)d

So 50^th term = 30 + ( 50-1)12 = ₹618

Example – 3:  The cost of borewell drilling cost per feet is ₹1000 for first feet and rises by ₹250 for each subsequent feet. Find the charge when good water found after digging borewell about 161 feet.

Solution: Here a = 1000 and d = 110 and nth term is 160

nth term = a + ( n-1)d

So charge for borewell work ( i.e 160 feet digging) = 1000 + ( 161-1)250 = ₹41,000

Arithmetic Progression Basic Problems

Example – 4:  Find the A.P with a  = -1.5 and d = -0.5

Solution: General form of A.M is a, a+d, a+2d, a+3d . . . . .

So A.M is -1.5, -2.0, -2.5, -3.0, -3.5 . . . . . . . .

Example – 5: Find the values of 20^th terms and sum of first 20 terms of the following series  1, 9, 17, 25, . . . . . . .

Solution: Here a = 1 ,  d=  8  

20^th term = a + ( n-1)d = 1 +  (19 x 8) = 153

sum of first 20 terms S20

S20 = ( 20/2) [ 2 x 1 + 19 x 8 ] = 1540

Example – 6:  Find the sum of first 16 terms of A.P 41, 36, 31, . . . . . .

Solution: Here a = 41, d = -5 and n= 16

Formula : 

S ₁₆ = ( 16/2) [ (2 x41) + (16-1) -5 ]

= 8 ( 7 ) = 56

Example – 7 :  In an A.P a ₇ = 30, then find sum of first 13 terms of that A.P

Solution: General form of A.P is a, a+d, a+2d, a+3d, . . . . . . . .

Here  a ₇ = 30 ;  so 7th term is a + 6d = 30 then   S ₁₃ = ?

S ₁₃ =  ( 13/2)   [ 2a + 12d ] = ( 13/2)  2  [ a + 6d ]  = 13 x 30 = 390

Example – 8: The sum of first 50 positive integers divisible by 3

Solution: Here the sequence is 3, 6, 9, 12, 15, . . . . .

So a = 3 , d = 3 and n = 50

S ₅₀ =  ( 50/2)   [ 2a + 49d ] =25 ( 2 x 3 + 49 x 3) = 25 x   153 =   = 3825

Example- 9: What will be the maximum sum of 42, 40, 38, 36 . . . . . . .

Solution: The maximum sum of the above sequence is “2” So

The series is 42 + 40 + 38 + . . . . . . + 2

Here a = 42 and d= 2 and n = 42/2 = 21 + 1 = 22

S₂₂ = (22/2) [ (2 x42) + (21 x -2)]  = 462

Example- 10: Find the 35^th term of the sequence of 3, 8, 9, 13, 15, 18, 21, 23, 27,  . . . . . .

Solution: The above sequence can be written as two sequences

i.e 3, 9, 15, 21, . . . . . . . .     & 8, 13, 18, 23, . . . . . . .

Now 35^th terms is equal to 18^th term of the 1^st sequence. So

a ₁₈ = 3 + (18-1)6 = 105

Arithmetic Progression Hard Questions

Example – 11:. The number of terms needed to get Sn = 0 in the A.P of 96, 93, 90,  . . . . . . . .

Solution: Here the sequence is 96, 93, 90,  . . . . . . . .

So a = 96 , d = – 3 and n = ? , S _n= 0 ;

S _n =  ( n/2)  [ 2 x 96 + (n-1)(- 3)] = 0

⇒ [ 192 – 3n + 3 ] = 0

⇒  [ 195 – 3n  ] = 0 ⇒ n = 65

So in the given  sequence 65 number of terms required

Example- 12: The n^th term of sequence of number is  a_n = n³ – 6n² + 11n – 6. Then find the sum of the first three terms of that sequence.

Solution: n^th term = n³ – 6n² + 11n – 6.

First three terms means n = 0, 1, & 2

Now substitute these values in above equation then -6, 0, 6.

So sum is -6+0+6 = 0

Example- 13: Find the Arithmetic progression if  a ₅ + a ₉  = 72  and a ₇ + a ₁₂ = 97.

Solution:  Here a ₅ + a ₉  = 72

⇒ ( a +4d) + (a + 8d) = 72

⇒ 2a + 12d = 72  – – – – -(  i )

And  a ₇ + a ₁₂ = 97.

⇒ ( a +6d) + (a + 11d) = 97

⇒ 2a + 17 d = 97  – – – – -(  ii )

From (i) and (ii)

a = 6  and d =5

So sequence is 6, 11, 16, 21, 26, . . . . . . .

Example- 14: In an A.P a_p = q an a_q = p then a_n = ?

Solution: Here a_p = q

i.e a_p =  a + ( p -1) d = q

⇒ a+ pd – d = q  – – – – -(  i )

And a_q = p

i.e a_q =  a + ( q -1) d = p

⇒ a+ qd – d = p  – – – – -(  ii )

From (i) and (ii)

d = -1 & a = p + q – 1

So a_n =  a + ( n-1)d = p + q – 1 + (n – 1) (-1) = p + q – n

Example- 15: In an A.P 31^st term is 40, then the sum of 61 terms of that A.P

Solution: Here 31^st terms is 40 So

a ₃₁   = 40

⇒ [ a +(31-1)d]  = 40

⇒ a + 30d = 40   – – – – -(  i )

S ₆₁ =?

⇒  S ₆₁ = ( 61/2) [2a + (61-1) d ] = 61  [ a + 30d]

Now substitute equation no. ( i) in above then

⇒  S ₆₁ =  61  [40]  = 2440

Example- 16: In an A.P  7^th and 21^st terms are 6 and -22 respectively. Find the 30^th term.

Solution: Here 7^th terms is 6  and 21^st term is -22 So

a ₇   = 6

⇒ [ a +(7-1)d]  = 6

⇒ a + 6d = 6   – – – – -(  i )

a ₂₁   = -22

⇒ [ a +(21-1)d]  = -22

⇒ a + 20d = -22   – – – – -(  ii )

From equations (i) and (ii)  a = 18 and d = -2

Now 30^th terms is

a ₃₀  = [ 18 + (29 x-2)] = -40

Example-17: Find the value of the expression 1 – 5 + 2 – 6 +3 -7 +4 -8 + . . . . . . . .  . to 100 terms

Solution: The given series is 1 – 5 + 2 – 6 +3 -7 +4 -8 + . . . . . . . .  . to 100 terms

The series can be rewritten as

⇒ (1 + 2 +3 +4 + . . . . . . . .  . to 50 terms) – (5 + 6 + 7 + 8 + . . . . . . . .  . to 50 terms)

Both these are in A.P

For 1st series  a=1 , d =1 & n = 50 , For 2nd series  a=5 , d =1 & n = 50 

=  (50/2) [ 2(1) + (49)(1) ] – (50/2) [ 2(5) + (49)(1) ]

= (25 x 51 ) – (25 x 59) = -200

Example-18: Find the sum of all number divisible by 7 in between  80 to 500.

Solution: Here 1st term = a = 84 ( which is the 1^st term greater than 80 that is divisible by 7.)

The last term less than 500, which is divisible by 7 is 497.

The number of terms in the in the AP ; 84, 112, 119, . . . . . . . .  497.

Here a = 84, d = 7  and number of terms = (497 – 84)/7 + 1 = 60

S_n = (60/2) [ (2 x 84) + (59 x 7) ] = 17430

Example – 19:  Find the sum of of first 200 terms of the following series 

1 + 4 + 6 + 5 + 11 + 6 + 16 + 7 +  . . . . . . . . . .  

Solution: The given series is 1 + 4 + 6 + 5 + 11 + 6 + 16 + 7 +  . . . . . . . . . .   ( 200 terms)

The above series treat every two consecutive terms as one.

i.e (1 + 4) +( 6 + 5) + (11 + 6) + (16 + 7) +  . . . . . . . . . .  ( 100 terms)

5 + 11 + 17 + 23 + . . . . . . .  ( 100 terms )

Here a = 5 ,  d = 6 and n = 100

S₁₀₀ = (100/2) [ (2 x 5) + (99 x 6)] = 30200

Example- 20; How many terms of the following series – 12, -9, -6, -3, . . . . . . .   must be taken the sum may be 54?

Solution: The given series is – 12, -9, -6, -3, . . . . . . . 

S_n = 54 , a = -12 ,  d = 3

54 = ( n /2) [ -24 + (n -1)3 ]

⇒ n² -9n – 36 = 0

By factorizing the above equation we can get n values of  12 & -3

Here the value of n cannot be negative. So n = 12

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