Arithmetic progressions - Problem based video part 4 for class 10th maths

1 Chapter : Arithmetic Progressions Website: www.letstute.com

Arithmetic Progressions

Problems based on Arithmetic Progressions

Given : Sequence = 23, 22 , 22, 21

Q) Which term of the sequence 23, 22 , 22, 21 … is the first negative term ?

12

12


Chapter : Arithmetic Progressions Website: www.letstute.com

To find: First negative term

Solution: a = first term = 23, 2nd term = 22 and

d = common difference = 22 - 23 = - 12

12

Then, an < 0

a + (n – 1)d < 0

23 + (n – 1) < 0



12

23 - + < 0n2

12

- < 047 2

n2

< 47 2

n2

47 < n

n > 47

Since, 48th is the natural number just greater than 47, therefore n = 48.

Thus, 48th term of the given sequence is the first negative term.



Q) If the mth term of an AP be and nth term be , then Show that its (mn)th term is 1.

1n

1m



Given: mth term =

nth term =

To prove: (mn)th term is 1

Solution: Let a = first term and d = common difference of the given AP.

am = a + (m – 1)d

= a + (m – 1)d am = , given … (1)1n

1n

and an = a + (n – 1)d



= a + (n – 1)d an = , given … (2) 1m

1m

Subtracting equation (2) from equation (1), we get

1n

1m

- = [a + (m – 1)d] - [a + (n – 1)d]

1n

- 1m = a + (m – 1)d - a – (n – 1)d

1n

- 1m

= (m – 1– n + 1)d

m - n mn

= (m– n)d



d = … (3) 1mn

Substituting the value of ‘d’ in equation (2), we get, 1 m = a + (n – 1) 1

mn

1m

- (n – 1) mn

= a

a = n – n + 1 mn

a = 1mn

….(4)



(mn)th term = amn = a + (mn – 1)d

amn = 1mn

[Using (3) and (4)]+ (mn – 1) 1mn

amn = 1mn

mn - 1 mn

+

amn = 1+ mn - 1 mn

mn mn

= = 1

Hence, the (mn)th term is 1



A negative term

from the given

sequence.

an = a + (n-1) d


Now we know…

Please visit www.letstute.com to take a test

Problems based onArithmetic Progressions

http://www.letstute.com/

Next video….

Some more problems based onArithmetic Progressions

Part 4

Please visit www.letstute.com to view the next video



Q) The sum of three consecutive numbers in AP is - 6, and their product is + 64. Find the numbers.



Given: S3 = - 6 Product of three consecutive numbers = 64

To Find: 3 consecutive numbers

Solution: Let the numbers be (a – d), a, (a + d)

Sum = - 6

(a – d) + a + (a + d) = - 6

a = - 2 …(1) 3a = - 6



(a – d) (a) (a + d) = + 64 a(a2 - d2) = + 64 -2[(-2)2 - d2] = + 64 [Using (1)] -2[4 - d2] = + 64 4 - d2 = - 32 d2 = 36

d = + 6



Product = + 64

If d = – 6, the numbers are [– 2 – (– 6)], – 2 and [– 2 + (– 6)], i.e, 4, –2 and – 8

Hence, the required numbers are – 8, – 2, 4 or 4, – 2, – 8.



If d = + 6, the numbers are (–2 – 6), – 2 and (– 2 + 6), i.e, – 8, –2 and 4

Q) Find the four consecutive even number of terms in AP whose sum is 16 and the sum of whose squares is 84.



Given: S4 = 16 Sum of squares of 4 consecutive even number of terms = 84

To Find: 4 consecutive even number of terms in AP

Solution: Let the four consecutive even number of terms in AP be a – 3d, a – d, a + d, a + 3d

Sum of the four consecutive even number of terms = 16

(a – 3d) + (a - d) + (a + d) + (a + 3d) = 16

a = 4 …(1)

4a = 16



a2 - 6ad + 9d2 + a2 - 2ad + d2 + a2 + 2ad + d2 + a2 +6ad + 9d2 = 84

(a – 3d)2 + (a - d)2 + (a + d)2 + (a + 3d)2 = 84

4a2 + 20d2 = 84

a2 + 5d2 = 21 42 + 5d2 = 21 [Using (1)]

5d2 = 5

d2 = 1

d = + 1



Sum of the squares of the four consecutive even number ofterms in AP = 84

4(a2 + 5d2) = 84

If d = + 1 then the terms are (4 – 3), (4 – 1), (4 + 1), (4 + 3) i.e. 1, 3, 5, 7.

If d = - 1 then the terms are (4 + 3), (4 + 1), (4 - 1), (4 - 3) i.e. 7, 5, 3, 1.

Hence, the required consecutive even number of terms are 1, 3, 5, 7 or 7, 5, 3, 1.



The AP may be written as

a, (a + d), (a + 2d),…. (l - 2d), (l – d), l

Last term from the end is l = l – (1 – 1)d

Second term from the end is l – d = l – (2 – 1)dThird term from the end is l – 2d = l – (3 – 1)d

Fourth term from the end is l – 3d = l – (4 – 1)d … and so on.

The nth term from the end is l – (n – 1)d



nth term from the end

Let the AP be a, a + d, a + 2d, ….. l

First term = a, common difference = d and last term = l

Q)Find the 12th term from the end of the AP 3, 6, 9,…60



Given: AP = 3, 6, 9,…60

To Find: 12th term from the end

Solution: d = common difference = 6 – 3 = 3 and the last term = l = 60

nth term from the end = l – (n – 1)d

= 60 – 11 x 3

Hence, the 12th term from the end of the AP 3, 6, 9… 60 is 27

12th term from the end = 60 – (12 – 1) x 3

= 60 – 33 = 27



Finding out exact numbers when the sum and product of numbers is given.

nth term from the end using the formula l – (n – 1)d

Treasure




Now we know…

Next slide - Problem based Video Please visit www.letstute.com to take a test

Problems based onArithmetic Progressions


Education

Arithmetic progressions - Problem based video part 4 for class 10th maths