[tex]\red {12}[/tex]
[tex] If \:a_{3} = 4,\:and\:a_{4} = 3\\in \:A.P \:then \:show\:that \:a_{7} = 0[/tex]
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[tex]\red {12}[/tex]
[tex] If \:a_{3} = 4,\:and\:a_{4} = 3\\in \:A.P \:then \:show\:that \:a_{7} = 0[/tex]
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[tex]\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}[/tex]
[tex]\Large \tt Given \begin{cases} \sf{a_3 \: = \: 4} \\ \sf{a_5 \: = \: 3} \end{cases}[/tex]
To Show :
a7 = 0
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Solution :
We know that Formula for terms is :
[tex]\LARGE {\underline{\boxed{\sf{a_n \: = \: a \: + \: (n \: - \: 1)d}}}}[/tex]
A.T.Q,
[tex]\Large \rightarrow {\sf{a_3 \: = \: a + (3 - 1)d}}[/tex]
Put Value of a3
[tex]\Large \rightarrow {\boxed{\sf{4 \: = \: a \: + \: 2d}}}---(1)[/tex]
And Similarly,
[tex]\Large \rightarrow {\sf{a_4 \: = \: a \: + \: (4 \: - \: 1)d}}[/tex]
[tex]\Large \rightarrow {\boxed{\sf{3 \: = \: a \: + \: 3d}}}----(2)[/tex]
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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
[tex]\LARGE \implies {\boxed{\boxed{\sf{a_7 \: = \: 0}}}}[/tex]
Hence Proved
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#answerwithquality
#BAL
Verified answer
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,
[tex]a_n=a+(n-1)d[/tex]
Now, Applying this formula for a₃ and a₄.
We have,
[tex]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[/tex]
Putting the value of d in equation i),
a = 4 - 2(-1)
a = 4 + 2
a = 6
Now,
[tex]a_7=a+(7-1)d\\\;\\a_7=6+6(-1)\\\;\\a_7=6-6\\\;\\a_7=0[/tex]