Arithmetic progressions - Problem no - 6 for class 10th maths

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

Arithmetic Progressions

Problems based on Arithmetic Progressions

Solution:

Q) The sum of first 9 terms of an AP is 351 and the sum of its first 20 terms is 1770. Find the first term of the AP and its common difference.

Sn= [2a + (n – 1)d] n 2

S9 = [2 x a + (9 – 1)d] 9 2

78 = 2a + 8d ….(1)

351 = (2a + 8d) [ S9 = 351 given] 9 2


Chapter : Arithmetic Progressions Website: www.letstute.com

351 x = 2a + 8d 2 9

and S20 = [2a + (20 – 1)d] 20 2

177 = 2a + 19d ….(2)

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

d = 9



1770 = 20[ 2a + (20 – 1)d] 2 1770 = 10 [2a +19d]

177 – 78 = [2a +19d] – [2a + 8d]

78 - 72 = 2a

Substituting the value of d in equation (1), we get

78 = 2a + 72

a = 3

Hence, the first term is 3 and the common difference is 9



78 = 2a + 8(9)

6 = 2a

Solution:

Q) The sum of first 10 terms of an AP is 485. If its last term is 71, find the first term and the common difference of the AP.

Let a = first term and d = common difference.

Sn = (a + l) n 2

Here, n = 10, l = 71 = a10 and S10 = 485

S10 = (a + 71) 10 2

485 = 5(a + 71) [ S10 = 485, given] ….(1) 97 = a + 71



∴

97 – 71 = a

a = 26

an = a + (n – 1)d a10 = 26 + (10 – 1)d [Using (1)]

71 = 26 + 9d [a10 = 71] 45 = 9d

d = 5

Hence, the first term is 26 and the common difference is 5



Solution:

Q)The first and last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Let a = first term and d = common difference of the AP

Here, a = 17, d = 9 and an = 350

an = a + (n – 1)d 350 = 17 + (n – 1)9 333 = 9n – 9 342 = 9n n = 38



Sn = (a + l) n 2

S38 = (17 + 350) 38 2

S38 = 19 (367)

S38 = 6973

Hence, there are 38 terms in the AP and their sum is 6973



Solution: a = first term = and d = common difference = 4 - =

Q) How many terms of the AP , 4, …. must beadded to obtain the sum 216 ?

11 3

13 3

11 3 11

3 1 3

Let the sum of ‘n’ terms be 216.

Sn = [2a + (n – 1)d]n2

216 = 2 x + (n – 1) n2

11 3

1 3

216 = + - n2

22 3

1 3

1n 3



216 = 7 + n2

n3

216 = + 7n 2

n2

6

1296 = 21n + n2

n2 + 21n – 1296 = 0



216 = 21n + n2

6

(n + 48) (n – 27) = 0

Either (n + 48) = 0 or (n – 27) = 0

n = -48 (rejected) or n = 27



n(n + 48) – 27(n + 48) = 0

Hence, the sum of 27 terms of the given AP is 216.

n2 + 48n – 27n 1296 = 0

n2 + 21n – 1296 = 0


Now we know…

Please visit www.letstute.com to take a test

Problems based onArithmetic Progressions

http://www.letstute.com/

Education

Arithmetic progressions - Problem no - 6 for class 10th maths