Question

is 68 a terms of AP7,9,10,13 yes or no​

Answer 1

Answer:

• No, 68 is not a term of given AP

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Step-by-step explanation:

In order to solve this problem, firstly we will assume that 68 is a term of the given AP and will try to find the position of 68 in the AP, i.e. value of n. If we succeed in finding the position of 68 in the given AP, will imply that it's a term of given AP. If not, then our answer would be "No"

Solution :-

Let's assume that 68 is a term of given AP.

We know,

$\bf\longrightarrow\:an\: =\: a \:+\: ( \:n \:- \:1 \:) \:d$

First term (a) of AP is 7 and it's common difference (d) is 2, also substitute an = 68

$\rm\longrightarrow\:an\: =\: a \:+\: ( \:n \:- \:1 \:) \:d$

$\rm\longrightarrow\:68\: =\: 7 \:+\: ( \:n \:- \:1 \:) \:(2)$

$\rm\longrightarrow\:68\:-\;7 =\: 2n \:- \:2$

$\rm\longrightarrow\:61\: =\: 2n \:- \:2$

$\rm\longrightarrow\:61\:+2\: =\: 2n$

$\rm\longrightarrow\:63\: =\: 2n$

$\rm\longrightarrow\:63\:/\:2 =\: n$

But since this value of n is coming out to be a decimal number but we know that any term of AP can only be expressed in the form of integer. From this we can conclude that 68 is not a term of given AP.