Question

Fifth term of an arithmetic sequence is 21 and its ninth term is 37.

what is its common difference?

what is first tearm?

explanation and answer please​

Answer 1

Answer:

The common difference 'd' and the first term 'a' of the AP are 4 and 5.

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Step-by-step explanation:

According to the question :

$\frak{ :\implies \: \: {5}^{th} \: term : a_{5} = 21 }$

$\frak{ : \implies \: \: {9}^{th } \: term : a_{9} =37 }$

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So, for this AP, let :

• Common difference be d
• First term be a
• Nᵗʰ term be n

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We know that :

$\frak{:\implies~~ a_n=a+(n-1)d}$

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Let's substitute the values for 5ᵗʰ term and 9ᵗʰ term :

(I) For a₅ :

$\frak{ : \implies \: \: a_{5} = a + (5 - 1)d}$

$\frak{ : \implies \: \:21 = a + 4d }$

$\frak{ : \implies ~~ 21-a=4d}$

$\frak{ :\implies~~ \dfrac{21-a}{4}=d~\dots\dots eq^n(i)}$

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(II) For a₉ :

$\frak{ : \implies \: \: a_{9} = a + (9 - 1)d }$

$\frak{: \implies \: \: 37 = a + 8d }$

$\frak {:\implies~~ 37-a=8d}$

$\frak{:\implies~~ \dfrac{37-a}{8}=d~\dots\dots eq^n(ii)}$

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We know that common difference 'd' will be same for both of them. So let's equate eqⁿ(i) and eqⁿ(ii) :

$\frak{:\implies~~\dfrac{21-a}{4}=\dfrac{37-a}{8}}$

$\frak{:\implies~~ \dfrac{21-a}{4}\times8=37-a}$

$\frak{:\implies~~ (21-a)\times2 =37-a}$

$\frak{:\implies~~ 42-2a=37-a}$

$\frak{:\implies~~ 42-37=-a+2a}$

$\frak{:\implies~~ 5=a}$

The first term 'a' is 5 for the given AP.

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Finally, substituting 'a' value in eqⁿ(i) to find 'd' :

$\frak{:\implies~~ \dfrac{21-a}{4}=d}$

$\frak{:\implies~~ \dfrac{21-5}{4}=d}$

$\frak{:\implies~~ \dfrac{16}{4}=d}$

$\frak{:\implies~~ 4=d}$

The common difference 'd' for the given AP is 4.