Math, asked by pattukonam, 1 month ago

I need step by step answers​

Attachments:

Answers

Answered by Anonymous
2

I HOPE THIS WILL HELP FUL FOR YOU

ANSWER 3.

n = 28

tn = 4n – 3

t1 = 4 x 1 – 3 = 1

t2 = 4 x 2 – 3 = 5

t28 = 4 x 28 – 3

= 112 – 3 = 109

∴ a = 1, d = t2 – t1 = 5 – 1 = 4

L = 109.

Sn = (n/2) (2a + (n – 1)d)

S28 = (28/2) (2 x 1 + 27 x 4)

= 14(2 + 108)

= 14 x 110

= 1540

Answer 4.

Let tn be nth term of an A.P.

tn = Sn – Sn - 1

= 2n2 – 3n – [2(n – 1)2 – 3(n – 1)] = 2n2 – 3n – [2(n2 – 2n + 1) – 3n + 3]

= 2n2 – 3n – [2n2 – 4n + 2 – 3n + 3]

= 2n2 – 3n – [2n2 – 7n + 5]

= 2n2 – 3n – 2n2 + 7n – 5

tn = 4n – 5

t1 = 4(1) – 5 = 4 – 5 = -1

t2 = 4(2) -5 = 8 – 5 = 3

t3 = 4(3) – 5 = 12 – 5 = 7

t4 = 4(4) – 5 = 16 – 5 = 11

The A.P. is -1, 3, 7, 11,… The common difference is 4

Ans. 5

given a 104= 125

a4=0

now.

to find- S35.

a + 103d=125

a+3d = 0

subtracting we get.

100d =125

d=125/100=5/4

now. d=5/4

a= -15/4..

S35= 35/2(2×-15/4 +34×5/4)

=35/2(-15/2+85/2)

= 35/2×35

=1225/2=612.5

therefore sum of first 35 terms is 612.5.

Ans 6.

Sum of odd positive integer less than 450

1 + 3 + 5 + …. 449

Here a = 1,

d = 3 – 1 = 2,

l = 449

Aliter: Sum of all the positive odd intergers = n2

= 225 × 225 = 50625

Sum of the odd integers less than 450 = 50625

Ans.7

The required sum

=(sum of all natural numbers from 603 to 901)-(sum of all natural numbers between 602 and 902 which are divisible by 4)

=224848-56400

=168448

The required sum

=(sum of all natural numbers from 603 to 901)-(sum of all natural numbers between 602 and 902 which are divisible by 4)

=224848-56400

=168448

Similar questions