Question

If the first three terms of an arithmetic sequence are $a,2a,18a-5b$ and further it is known that $a+b+4=0,$ then the value of the $100th$ term of the sequence would be?

• A. $-100$
• B. $-300$
• C. $150$
• D. $-150$

Answer 1

The correct option is A $-100$

Since it is given that the first three terms of in arithmetic sequence, therefore, the difference between the consecutive terms remains constant and same. The first three terms are given as follows a, 2a and 18a-5b. Thus second term - first term = third term - second term that is:

$2a-a=18a-5b-2a$

$a=16a-5b$

$15a=5b$

$3a=b$

Also it is given that a+b+4=0, thus

$a+b+4=0$

$a+3a+4=0$

$4a=-4$$

$a=-1$$

$\Rightarrow$$3a=b$ will become

$\Rightarrow$$b=-3$

Therefore, the first three terms are as follows:

-1, -2, -3, .....

Hence the 100th term is

=$a+(n-1)d$

=$a+(100-1)d$

=$a+\left(99\right)d$

=$-1-1\left(99\right)$

=$-1-99$

=$-100$

Hence option A is correct.