Question

2,5,9,14,....... is an A.P.
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Answer 1

$\huge{\sf{=====Answer=====}}$

In which :-

$\large{\sf{a_{1} \: = \: 2 }}$

$\large{\sf{ a_{2} \: = \: 5}}$

$\large{\sf{a_{3} \: = \: 9}}$

$\large{\sf{a_{4} \: = \: 14}}$

We need to see common Difference for checking whether it is an A.P. or not .

$\large{\sf{D \: = \: a_{2} \: - \: a_{1}}}$

__________________[Put Values]

$\large{\sf{= \: 5 \: - \: 2}}$

$\large{\implies}{\boxed{\sf{ D \: = \: 3}}}$

$\rule{200}{2}$

Similarly,

$\large{\sf{D \: = \: a_{3} \: - \: a_{2}}}$

__________________[Put Values]

$\large{\sf{ = \: 9 \: - \: 5}}$

$\large{\implies}{\boxed{\sf{D \: = \: 4}}}$

a2 - a1 ≠ a3 - a2

And

So, It is not a A.P. Because Common difference is not same.

✌ ✌

Answer 2

Answer:

Given sequence is not an A.P

Step-by-step explanation:

/* An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term */

Given sequence 2,5,9,14, ....

$a_{2}-a_{1}= 5-2 = 3$

$a_{3}-a_{2}= 9-5 = 4$

$a_{2}-a_{1}≠a_{3}-a_{2}$

Therefore,

Given sequence is not an A.P

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