# Maths homework help!?

+1 vote
1. Find the first three terms of the nth term of the APs in which:
The sum of the 4th and 8th terms is 10, and the sums of the 6th and 12th terms is 22

2.Find the sum of all the multiples of 3 and 0 between 100 using APs

3a. How many numbers between 100 and 200 are divisible by 7?
b. Find their sum

1). Arithmetic Sequence in nth term is [a + (n-1)d]

The sum of the 4th and 8th terms is 10, and the sums of the 6th and 12th terms is 22

i.e. 4th term + 8th term = 10 and 6th term + 12th term = 22.

So, 12th term = 10 and 18th term = 22.

a + 11d = 10------------->(1)

a + 17d = 22------------->(2)

Solve the equation (1) and equation (2).

a + 11d = 10.

a + 17d = 22.

--------------------------

- 6d = - 12

Divide each side by - 6.

d = 2.

Substituting d = 2 in the equation (1)

a + 11(2) = 10

a + 22 = 10

Subtract 22 from each side.

a = - 12.

The first three terms in the APs is a, a + d, a + 2d.

a = -12,

a + d = -12 + 2 = - 10,

a  + 2d = -12 + 2(2) = -12 + 4 = - 8.

The first three terms in the APs is -12, -10, -8, ---------------

The first three terms of the n th term of the APs are - 5, - 3, and - 1.

+1 vote

2). The sum of all the multiples of 3 and 0 between 100 using APs

i.e. 3 + 6 + 9 + ----------- + 99.

The sum of n terms formula: Sn = (n/2)(a + l)

Here a = 3, l = 99 and n = number of terms = 33

Sn = (33/2)(3 + 99)

= (33/2)(102)

= 33(51)

= 1683.

Therefore The sum of all the multiples of 3 and 0 between 100  is 1683.

+1 vote

3a). Multiples of 7 between 100 and 200 are 105, 112, 119, ----------, 196.

So, the total numbers are 14.

b). Sum is 105 + 112 + 119 + --------------- + 196.

This is arithmetic series so Sn = (n/2)(a + l)

Here a = 105, l = 196 and n = 14.

Sn = (14/2)(105 + 196)

Sn = 7(301)

Sn = 2107.

1).

In arithmetic Sequence, the nth term is [a + (n-1)d].

From the given data : 4 th term + 8 th term = 10.

Means that, a + 3d + a + 7d = 10

2a + 10d = 10

⇒ a + 5d = 5 → (1), and

6 th term + 12 th term = 22.

Means that, a + 5d + a + 11d = 22

2a + 16d = 22

⇒ a + 8d = 11 → (2).

Solve the equations (1) and (2).

a + 5d = 5

a + 8d = 11

(-)_____________

- 3d = - 6

⇒ d = 2.

Substitute, d = 2 in equation (1).

a + 5(2) = 5

a = 5 - 10

⇒ a = - 5.

The first three terms of the n th term of the APs : a, a + d, and a + 2d = - 5, (- 5 + 2), and (- 5 + 2*2) = - 5, - 3, and - 1.