Question

Let a1,a2,a3.... be an A.P, such that a1 + a2 + ...+ap/a1+a2+a3 ....+aq =p^3/q^3 . Then a6/a21 is equal to
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Answer 1

Answer:

3

Step-by-step explanation:

If (a1+a2+... ap.)/(a1+a2+... aq) = p2/q2, p not equal to q, then a6/a21 equals (1) 7/2 (2) 2/7 (3) 11/41 (4) 41/11. Hence option (3) is the answer.

Answer 2

$\large\underline{\sf{Solution-}}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Now, given that,

$\rm :\longmapsto\:\dfrac{a_1 + a_2 + a_3 + - - - + a_p}{a_1 + a_2 + a_3 + - - - + a_q} = \dfrac{ {p}^{3} }{ {q}^{3} }$

Let assume that First term is represented by a and common difference of the series is d.

So,

$\rm :\longmapsto\:\dfrac{\dfrac{p}{2} \bigg(2 \:a\:+\:(p\:-\:1)\:d \bigg)}{\dfrac{q}{2} \bigg(2 \:a\:+\:(q\:-\:1)\:d \bigg)} = \dfrac{ {p}^{3} }{ {q}^{3} }$

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+\:(p\:-\:1)\:d}{ 2 \:a\:+\:(q\:-\:1)\:d } = \dfrac{ {p}^{2} }{ {q}^{2} }$

Now, Substitute p = 11 and q = 41, so we get

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+\:(11\:-\:1)\:d}{ 2 \:a\:+\:(41\:-\:1)\:d } = \dfrac{ {11}^{2} }{ {41}^{2} }$

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+\:(10)\:d}{ 2 \:a\:+\:(40)\:d } = \dfrac{121}{ 1681 }$

$\rm :\longmapsto\:\dfrac{ 2 (\:a\:+\:5\:d)}{ 2( \:a\:+ \: 20\:d )} = \dfrac{121}{ 1681 }$

$\rm :\longmapsto\:\dfrac{ \:a\:+\:5\:d}{ \:a\:+ \: 20\:d } = \dfrac{121}{ 1681 }$

$\bf\implies \:\:\dfrac{ a_6}{ a_{21} } = \dfrac{121}{ 1681 }$

Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.