Question

common difference of an arithmetic sequence is 7 and 11nth term is 80.
a) which term of the sequence is 185?
b)can 2019 be a term of the sequence? Justify

Answer 1

Step-by-step explanation:

1. 26
2. No

Sorry guy i have no time to solve i did it with calculator

Answer 2

a) 26th term of the sequence is 185.

b) Yes, 2019 can be a term of sequence.

Step-by-step explanation:

We are given with a AP whose common difference (d) is 7 and 11th term(a11) is 80. By using given data we can write,

$d=7$ and $a11=80$.

we have to find the value of n where sequence is 185 and check 2019 can be a term of this sequence or not.

• Formula used,

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

where 'a' is the first term of sequence, 'd' is the common distance of AP and 'n' is the term number.

• Calculation for 'a'

we are given the 11th term of AP and common distance, so by this given data we can find the value of first term of AP 'a',

$d=7$

$a11=80$

According to formula,

$a11=a+(11-1)*7$

$80=a+10*7$

$80=a+70$

$a=80-70$

$a=10$

We get the first term of AP which is 10.

• Calculation for part (a)

In this part we have the value of an but not know the value of n so we have to find the n. By using formula we can write,

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

$an=185$  , $a=10$  and  $d=7$

$185=10+(n-1)*7$

Opening bracket by doing multiplication,

$185=10+7*n-7*1$

$185=10+7n-7$

$185=7n+3$

$7n=185-3$

$7n=182$

$n=\frac{182}{7}$

$n=26$

So the 185 is the 26th term of sequence.

• Calculation for part (b)

We have to check 2019 can be a term of given AP or not. So for this, we use same formula and find the value of 'n' same as we done in part (a).

Now the value of an is different which is 2019.

$an=2019$  , $a=10$  and  $d=7$

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

$2019=10+(n-1)*7$

$2019=10+7n-7$

$2019=7n+3$

$7n=2019-3$

$7n=2016$

$n=\frac{2016}{7}$

$n=288$

2019 is the 288th term of sequence so it means 2019 is the term of sequence.