# 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
asked Feb 21, 2013

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, ---------------

answered Feb 21, 2013

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.

answered Feb 21, 2013
+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.

answered Feb 21, 2013

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.

answered Jul 28, 2014