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# RD Chapter 19- Arithmetic Progressions Ex-19.3 Interview Questions Answers

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Question 1 : The Sum of the three terms of an A.P. is 21 and the product of the first, and the third terms exceed the second term by 6, find three terms.

Given:

The sum of first threeterms is 21

Let us assume thefirst three terms as a – d, a, a + d [where a is the first term and d is thecommon difference]

So, sum of first threeterms is

a – d + a + a + d = 21

3a = 21

a = 7

It is also given thatproduct of first and third term exceeds the second by 6

So, (a – d)(a + d) – a= 6

a2 – d2 –a = 6

Substituting the valueof a = 7, we get

72 – d2 –7 = 6

d2 =36

d = 6 or d = – 6

Hence, the terms of APare a – d, a, a + d which is 1, 7, 13.

Question 2 : Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers

Given:

Sum of first threeterms is 27

Let us assume thefirst three terms as a – d, a, a + d [where a is the first term and d is thecommon difference]

So, sum of first threeterms is

a – d + a + a + d = 27

3a = 27

a = 9

It is given that theproduct of three terms is 648

Substituting the valueof a = 9, we get

93 –9d2 = 648

729 – 9d2 =648

81 = 9d2

d = 3 or d = – 3

Hence, the given termsare a – d, a, a + d which is 6, 9, 12.

Question 3 : Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Given:

Sum of four terms is50.

Let us assume thesefour terms as a – 3d, a – d, a + d, a + 3d

It is given that, sumof these terms is 4a = 50

So, a = 50/4

= 25/2 … (i)

It is also given thatthe greatest number is 4 time the least

a + 3d = 4(a – 3d)

Substitute the valueof a = 25/2, we get

(25+6d)/2 = 50 – 12d

30d = 75

d = 75/30

= 25/10

= 5/2 … (ii)

Hence, the terms of APare a – 3d, a – d, a + d, a + 3d which is 5, 10, 15, 20

Question 4 : The sum of three numbers in A.P. is 12, andthe sum of their cubes is 288. Find the numbers.

Given:

The sum of threenumbers is 12

Let us assume thenumbers in AP are a – d, a, a + d

So,

3a = 12

a = 4

It is also given thatthe sum of their cube is 288

(a – d)3 +a3 + (a + d)3 = 288

a3 – d3 –3ad(a – d) + a3 + a3 + d3 +3ad(a + d) = 288

Substitute the valueof a = 4, we get

64 – d3 –12d(4 – d) + 64 + 64 + d3 + 12d(4 + d) = 288

192 + 24d2 =288

d = 2 or d = – 2

Hence, the numbers area – d, a, a + d which is 2, 4, 6 or 6, 4, 2

Question 5 :

If the sum of three numbers in A.P. is 24 and their product is 440, findthe numbers.

Given:

Sum of first threeterms is 24

Let us assume thefirst three terms are a – d, a, a + d [where a is the first term and d is thecommon difference]

So, sum of first threeterms is a – d + a + a + d = 24

3a = 24

a = 8

It is given that theproduct of three terms is 440

Substitute the valueof a = 8, we get

83 –8d2 = 440

512 – 8d2 =440

72 = 8d2

d = 3 or d = – 3

Hence, the given termsare a – d, a, a + d which is 5, 8, 11

Question 6 :

The angles of a quadrilateral are in A.P. whose common difference is 10.Find the angles

Given: d = 10

We know that the sumof all angles in a quadrilateral is 360

Let us assume theangles are a – 3d, a – d, a + d, a + 3d

So, a – 2d + a – d + a+ d + a + 2d = 360

4a = 360

a = 90… (i)

And,

(a – d) – (a – 3d) =10

2d = 10

d = 10/2

= 5

Hence, the angles area – 3d, a – d, a + d, a + 3d which is 75o, 85o, 95o,105o

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