Question

the fourth term of an arithmetic requence is 10 and the sixth term is 16 determine the sequence?
a. 2, 4,6,8,10....
b. 3,4,5,6,7.....
c. 1,4,7,10,13....
d. 2,4,5,6,10....​

Answer 1

Step-by-step explanation:

option c is the correct

Answer 2

The sequence of AP is (c) 1, 4, 7, 10, 13.

Step-by-step explanation:

We are given the fourth and sixth term of an AP which are 10 and 16 respectively such as,

$a4=10$

$a6=16$

we have to find the sequence and for determining the sequence we have to find the first term(a) and common difference(d) of AP.

• Formula used,

$an=a+(n-1)*d$

where a is the first term of AP, d is the common difference and n is the term number of AP.

• Calculation for 'a' and 'd'

we have the 4th and 6th term so by using formula we can write them as ,

$a4=10$   here $n=4$,

$a4=a+(4-1)*d$

$a+3d=10$        

$a6=16$   here $n=6$,

$a6=a+(6-1)*d$

$a+5d=16$      

So we get two linear equations ,

$a+3d=10$        ---(1)

$a+5d=16$       ---(2)

now we can solve these linear equations by using elimination method.

first extract a from equation (1)

$a+3d=10$  

$a=10-3d$

then put this a in equation (2),

$a+5d=16$

$10-3d+5d=16$

taking 10 to other side

$-3d+5d=16-10$

$2d=6$

$d=\frac{6}{2}$

$d=3$

put this value of d in extracted 'a'

$a=10-3d$

$a=10-3*3$

  $=10-9$

$a=1$

In this way we get first term and common difference of AP as,

$a=1$   and  $d=3$.

• Determination of Sequence

First term is $a=1$

by using formula the second term is,

$a2=a+(2-1)*d$

    $=1+1*3$

$a2=4$

similarly $a3=7$ ,  $a4=10$   and $a5=13$.

so the sequence is $1,4,7,10,13..$

Correct option is option (c).