Question

If the ratio of the fifth and tenth term of an A.P is 1:2, then find the ratio of the first term to the common difference of the given A.P?

Answer 1

Answer:

Step-by-step explanation:

let a be the first term and d be the common difference.

We know that nth term of AP, tₙ = a + (n - 1)d

5th term, t₅ = a + (5 - 1)d = a + 4d

10th term, t₁₀ = a + (10  - 1)d = a + 9d

given t₅ : t₁₀ = 1 : 2

=> a + 4d/a + 9d = 1/2

//by cross multiplication,

=> 2( a + 4d) = 1(a + 9d)

=> 2a + 8d = a + 9d

=> a = d

=> a/d = 1

Thus ratio of first term to common difference is 1: 1

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• Ratio of the fifth and tenth term of an A.P is 1:2

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$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• Ratio of the first term to the common difference of the given A.P

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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$\underline{\bold{\texttt{We know that :}}}$

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

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• $\sf a_n$ = nth term

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• a = First term

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• n = Number of term

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• d = Common difference

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$\underline{\bold{\texttt{5th term of AP }}}$

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➜ $\sf a_5 = a + (5 - 1)d$

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➜ $\sf a_5 = a + 4d$

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$\underline{\bold{\texttt{10th term of AP }}}$

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➜ $\sf a_{10} = a + (10 - 1)d$

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➜ $\sf a_{10} = a + 9d$

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〚 Ratio of the fifth and tenth term of an A.P is 1:2 〛

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So,

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➜ $\sf \dfrac { a_5 } { a_{10} } = \dfrac { 1 } { 2 }$

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Putting the values

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➜ $\sf \dfrac {a + 4d } {a + 9d} = \dfrac { 1 } { 2 }$

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〚 Cross multiplying it 〛

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➜ a + 9d = 2a + 8d

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➜ 9d - 8d = 2a - a

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➜ d = a ----- (1)

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〚 Now we need to evaluate the ratio of the first term to the common difference of the given A.P 〛

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➜ $\sf \dfrac { a } { d }$ ---- (2)

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〚 But from (1) we got that d = a 〛

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So, equation (2) can also be written as ,

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➜ $\sf \dfrac { d } { d }$

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➜ $\sf \dfrac {\cancel d } { \cancel d }$

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➨ $\sf \dfrac { a } { d } = \dfrac { 1 } { 1 }$

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• Therefore ratio of the first term to the common difference of the given A.P is 1:1