## RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
- RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

**Question 1.**

Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P.

**Solution:**

(8x + 4), (6x – 2) and (2x + 7) are in A.P.

(6x – 2) – (8x + 4) = (2x + 7) – (6x – 2)

=> 6x – 2 – 8x – 4 = 2x + 7 – 6x + 2

=> -2x – 6 = -4x + 9

=> -2x + 4x = 9 + 6

=> 2x = 15

Hence x = \(\frac { 15 }{ 2 }\)

**Question 2.**

If x + 1, 3x and 4x + 2 are in A.P., find the value of x.

**Solution:**

x + 1, 3x and 4x + 2 are in A.P.

3x – x – 1 = 4x + 2 – 3x

=> 2x – 1 = x + 2

=> 2x – x = 2 + 1

=> x = 3

Hence x = 3

**Question 3.**

Show that (a – b)², (a² + b²) and (a + b)² are in A.P.

**Solution:**

(a – b)², (a² + b²) and (a + b)² are in A.P.

If 2 (a² + b²) = (a – b)² + (a + b)²

If 2 (a² + b²) = a² + b² – 2ab + a² + b² + 2ab

If 2 (a² + b²) = 2a² + 2b² = 2 (a² + b²)

Which is true

Hence proved.

**Question 4.**

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

**Solution:**

Let the three terms of an A.P. be a – d, a, a + d

Sum of three terms = 21

=> a – d + a + a + d = 21

=> 3a = 21

=> a = 7

and product of the first and 3rd = 2nd term + 6

=> (a – d) (a + d) = a + 6

a² – d² = a + 6

=> (7 )² – d² = 7 + 6

=> 49 – d² = 13

=> d² = 49 – 13 = 36

=> d² = (6)²

=> d = 6

Terms are 7 – 6, 7, 7 + 6 => 1, 7, 13

**Question 5.**

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

**Solution:**

Let the three numbers of an A.P. be a – d, a, a + d

According to the conditions,

Sum of these numbers = 27

a – d + a + a + d = 27

=> 3a = 27

**Question 6.**

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

**Solution:**

Let the four terms of an A.P. be (a – 3d), (a – d), (a + d) and (a + 3d)

Now according to the condition,

Sum of these terms = 50

=> (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50

=> a – 3d + a – d + a + d + a – 3d= 50

=> 4a = 50

=> a = \(\frac { 25 }{ 2 }\)

and greatest number = 4 x least number

=> a + 3d = 4 (a – 3d)

=> a + 3d = 4a – 12d

=> 4a – a = 3d + 12d

**Question 7.**

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

**Solution:**

**Question 8.**

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6. **[CBSE 2016]**

**Solution:**

**Question 9.**

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

**Solution:**

Let the four angles of a quadrilateral which are in A.P., be

a – 3d, a – d, a + d, a + 3d

Common difference = 10°

Now sum of angles of a quadrilateral = 360°

a – 3d + a – d + a + d + a + 3d = 360°

=> 4a = 360°

=> a = 90°

and common difference = (a – d) – (a – 3d) = a – d – a + 3d = 2d

2d = 10°

=> d = 5°

Angles will be

a – 3d = 90° – 3 x 5° = 90° – 15° = 75°

a – d= 90° – 5° = 85°

a + d = 90° + 5° = 95°

and a + 3d = 90° + 3 x 5° = 90° + 15°= 105°

Hence the angles of the quadrilateral will be

75°, 85°, 95° and 105°

**Question 10.**

Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623. **[NCERT Exemplar]**

**Solution:**

Let the three parts of the number 207 are (a – d), a and (a + d), which are in A.P.

Now, by given condition,

=> Sum of these parts = 207

=> a – d + a + a + d = 207

=> 3a = 207

a = 69

Given that, product of the two smaller parts = 4623

=> a (a – d) = 4623

=> 69 (69 – d) = 4623

=> 69 – d = 67

=> d = 69 – 67 = 2

So, first part = a – d = 69 – 2 = 67,

Second part = a = 69

and third part = a + d = 69 + 2 = 71

Hence, required three parts are 67, 69, 71.

**Question 11.**

The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles. **[NCERT Exemplar]**

**Solution:**

Given that, the angles of a triangle are in A.P.

**Question 12.**

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number. **[NCERT Exemplar]**

**Solution:**

or, d = ± 2

So, when a = 8, d = 2,

the numbers are 2, 6, 10, 14.

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5 are helpful to complete your math homework.

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