Question

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$If \:a_{3} = 4,\:and\:a_{4} = 3\\in \:A.P \:then \:show\:that \:a_{7} = 0$​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\Large \tt Given \begin{cases} \sf{a_3 \: = \: 4} \\ \sf{a_5 \: = \: 3} \end{cases}$

To Show :

a7 = 0

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Solution :

We know that Formula for terms is :

$\LARGE {\underline{\boxed{\sf{a_n \: = \: a \: + \: (n \: - \: 1)d}}}}$

A.T.Q,

$\Large \rightarrow {\sf{a_3 \: = \: a + (3 - 1)d}}$

Put Value of a3

$\Large \rightarrow {\boxed{\sf{4 \: = \: a \: + \: 2d}}}---(1)$

And Similarly,

$\Large \rightarrow {\sf{a_4 \: = \: a \: + \: (4 \: - \: 1)d}}$

$\Large \rightarrow {\boxed{\sf{3 \: = \: a \: + \: 3d}}}----(2)$

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Solve (1) and (2)

⇒4 = a + 2d

a = 4 - 2d

Put value of a in (2)

3 = 4 - 2d + 3d

3 - 4 = - 2d + 3d

⇒-1 = d

Value of d is -1 .

Now put value of a in (1)

4 = a + 2d

⇒a + 2(-1) = 4

⇒a - 2 = 4

⇒a = 4 + 2

⇒a = 6

Now put value of a and d in the formula for an

⇒a7 = 6 + (7 - 1)(-1)

⇒a7 = 6 + 6(-1)

⇒a7 = 6 - 6

⇒a7 = 0

$\LARGE \implies {\boxed{\boxed{\sf{a_7 \: = \: 0}}}}$

Hence Proved

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#answerwithquality

#BAL

Answer 2

Answer: a₇ = 0

Step-by-step explanation:

Given that,

a₃ = 4

a₄ = 3

Let the first term of AP be a with common difference d in the terms.

We know that nth term of AP is given by the formula,

$a_n=a+(n-1)d$

Now, Applying this formula for a₃ and a₄.

We have,

$a_3=a+(3-1)d\\\;\\4=a+2d\\\;\\a=4-2d\;\;\;\;..............i)\\\;\\\textbf{and,}\\\;\\a_4=a+(4-1)d\\\;\\3=a+3d\\\;\\\text{Putting the value of a from equation i) in this equation,}\\\;\\3=4-2d+3d\\\;\\3-4=d\\\;\\d=-1$

Putting the value of d in equation i),

a = 4 - 2(-1)

a = 4 + 2

a = 6

Now,

$a_7=a+(7-1)d\\\;\\a_7=6+6(-1)\\\;\\a_7=6-6\\\;\\a_7=0$