Question

8th term of an arithmetic sequence is 67 and 18th term is 147. Is 125 a term of this sequence?
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Answer 1

Answer:

No, 125 is not a term of the sequence whose 8th term is 67 and 18th term is 147.

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Answer 2

Given :

• 8th term of an AP is 67
• 18th term of an AP is 147

To Find :

125 is a term of this sequence or not

Formula's :

• Genral term of an AP

$\sf\purple{a_n=a+(n-1)d}$

Where ,

• a = first term of an A.P
• d = common difference
• n = no of terms

Solution:

Given : 8th term of an AP is 67

$\sf\:a_8=a+(8-1)d$

$\sf\implies\:a_8=a+7d$

$\sf\implies\:a+7d=67...(1)$

and,18th term of an AP is 147

$\sf\:a_{18}=a+(18-1)d$

$\sf\implies\:a_18=a+17d$

$\sf\implies\:a+17d=147$

$\sf\implies\:a+7d+10d=147$

Now , From equation (1)

$\sf\implies\:67+10d=147$

$\sf\implies10d=147-67$

$\sf\implies10d=80$

$\sf\implies\:d=8$

Put the value of d = 8 in equation (1) then ,

$\sf\:a+7(8)=67$

$\sf\implies\:a+56=67$

$\sf\implies\:a=67-56$

$\sf\implies\:a=11$

Thus , a = 11 and d = 8

We have to find 125 is a term of this sequence or not .

Let 125 be $\sf\:a_n$ term of the given sequence . then ,

$\sf125=a+(n-1)d$

$\sf\implies125=11+(n-1)(8)$

$\sf\implies125-11=(n-1)(8)$

$\sf\implies\:(n-1)=\dfrac{114}{8}$

$\sf\implies\:n=\dfrac{114}{8}+1$

$\sf\implies\:n=\dfrac{122}{8}$

$\sf\implies\:n=15.25$

Since , No of terms can't be negative .Hence, 125 is not a term of given sequence.