Question

which of the following is not A.p,. a) 2,4,5,6...... ,. b)2,5,8,8,..... c)1,11,21,..... d)6,5,4,3,..​

Answer 1

Here, 1st and 2nd sequence are not in ap bcoz their common differences are not constant
And 3rd, 4th sequences are in ap bcoz they have common constant common differences which are 10 and -1 respectively
..........mark it as brainliest-...............

Answer 2

Question:-

☞Which of the following is not AP.

(a) 2,4,5,6 ...

(b)2,5,8,8....

(c)1,11,21...

(d) 6,5,4,3....

Answer:-

$\implies$ Option (a)

$\implies$Option(b)

Solution:-

$\implies$ Firstly to know that the following term is not an AP we have to know the common diffrence(d) . if the common diffrence is not same so the following will not be an AP.

$\sf {\tt {\boxed {\red{(a) 2,4,5,6....}}}}$

$\implies HERE, 2=a_{1} , 4=a_{2}, 5=a_{3}, 6=a_{4} \\\\ \implies a_{4} - a_{3} = 6 -5 = 1 \\\\ \implies a_{3} - a_{2} = 5 - 4 = 1 \\\\ \implies a_{2} - a_{1} = 4 - 2 = 2$

So. here 1, 1, 2 is the diffrence which is not same so option (a) is not an AP.

$\sf {\tt {\boxed {\pink {(b) 2,5,8,8....}}}}$

$\implies HERE, 2=a_{1} , 5=a_{2} , 8=a_{3}, 8=a_{4} \\\\ \implies a_{4} - a_{3} = 8 - 8= 0 \\\\ \implies a_{3} - a_{2} = 8 - 5= 2 \\\\ \implies a_{2} - a_{1} = 5 - 2= 3$

So, here 0, 2, 3 is the diffrence which is not same so option (b) is not an AP

$\sf{\tt {\boxed {\red {(c) 1, 11, 21......}}}}$

$\implies HERE, 1=a_{1} , 11=a_{2}, 21=a_{3} ,\\\\ \implies a_{3} - a_{2} = 21-11= 10 \\\\ \implies a_{2} - a_{1} = 11 - 1= 10$

So, here the common diffrence is same so it is AP

$\sf{\tt {\pink {\boxed {(d)6,5,4,3.....}}}}$

$\implies HERE, 6=a_{1} , 5=a_{2} , 4=a_{3}, 5=a_{4} \\\\ \implies a_{4} - a_{3} = 3-4= -1 \\\\ \implies a_{3} - a_{2} = 4 -5 = -1\\\\ \implies a_{2} - a_{1} = 5 -6 = -1$

So, here common diffrenece is same so it is an AP.

$\star$ hence, option(a) , (b), is not an AP.