Question

Find the first term and common difference of an ap.if the 4th term is 14 and the eleventh term is 42

Answer 1

Explanation:

an arithmetic sequence is always built starting with a given a1 (first term) and a constant number d (the common difference) that is added for every new term.

so,

a1 = a1

a2 = a1 + d

a3 = a2 + d = a1 + d + d = a1 + 2d

a4 = a3 + d = a2 + d + d = a1 + d + d + d = a1 + 3d

...

you see the structure ?

an = an-1 + d = a1 + (n-1)×d

we have now 2 unknown numbers (a1 and d) and need therefore 2 terms to create 2 equations with the 2 variables and solve them.

luckily we got 2 : a4 and a11

a4 = 14 = a1 + 3d

a11 = 42 = a1 + (11-1)×d = a1 + 10d

summary of the 2 equations :

14 = a1 + 3d

42 = a1 + 10d

we can now subtract the second equation from the first to eliminate a1, and the we solve the result for d.

at the end we use then one of the original equations with the discovered d to solve for a1.

14 = a1 + 3d

- 42 = a1 + 10d

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

-28 = 0 - 7d

28 = 7d

d = 28/7 = 4

let's pick e.g. the first equation :

14 = a1 + 3×4 = a1 + 12

a1 = 2

Answer 2

The first term of the arithmetic progression is 2, and the common difference is 4. This is determined by the given information that the 4th term is 14 and the 11th term is 42.

In an arithmetic progression (AP), each term is obtained by adding a constant difference to the preceding term. The general form of an arithmetic sequence is given by:

`\[ a_n = a_1 + (n-1)d \]`

where:

-
`\( a_n \)` is the
`\(n\)-th` term,

-
`\( a_1 \)` is the first term,

-
`\( d \)` is the common difference,

-
`\( n \)` is the term number.

Given that the 4th term
`(\( a_4 \))` is 14 and the 11th term (
`\( a_(11) \))` is 42, we can set up two equations:

1. For the 4th term:

`\[ a_4 = a_1 + 3d = 14 \]`

2. For the 11th term:

`\[ a_(11) = a_1 + 10d = 42 \]`

Now, we can solve these two equations simultaneously to find
`\( a_1 \)`and
`\( d \).`

From the first equation:

`\[ a_1 + 3d = 14 \]`

From the second equation:

`\[ a_1 + 10d = 42 \]`

Subtract the first equation from the second to eliminate
`\( a_1 \):`

`\[ 7d = 28 \]`

Now, solve for \( d \):

`\[ d = (28)/(7) = 4 \]`

Now that we have
`\( d \),` substitute it back into the first equation to find
`\( a_1 \):`

`\[ a_1 + 3(4) = 14 \]`

`\[ a_1 + 12 = 14 \]`

`\[ a_1 = 2 \]`

So, the first term
`(\( a_1 \))` is 2, and the common difference
`(\( d \))` is 4.