Question

Kindly solve this 29 question, options are given.
Given correct answer along with explanation.​

Answer 1

Answer:

option no- c-2/7

Step-by-step explanation:

6 divided by 2

21 divided by 7

there four answer is 2/7

Answer 2

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

Given that,

$\rm :\longmapsto\:a_1,a_2,a_3, - - - - ,a_n \: are \: in \: AP$

Let assume that

↝ First term of an AP is a

↝ Common difference of an AP is d

Now,

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, Consider,

$\red{\rm :\longmapsto\:a_1 + a_2 + a_3 + - - - - + a_p \: }$

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

Now, Consider,

$\red{\rm :\longmapsto\:a_1 + a_2 + a_3 + - - - - + a_q \: }$

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

Now, Given that

$\rm :\longmapsto\:\dfrac{a_1 + a_2 + a_3 + - - - - + a_p}{a_1 + a_2 + a_3 + - - - - + a_q} = \dfrac{ {p}^{2} }{ {q}^{2} }$

So, on substituting the values, evaluated above, we get

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

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

$\rm :\longmapsto\:\dfrac{2a + dp - d }{2a + dq -d} = \dfrac{ {p} }{ {q}}$

$\rm :\longmapsto\:2aq + dpq - dq = 2ap + dpq - dp$

$\rm :\longmapsto\:2aq - dq = 2ap - dp$

$\rm :\longmapsto\:2aq - 2ap =dq - dp$

$\rm :\longmapsto\:2a(q - p)=d(q - p)$

$\bf\implies \:d \: = \: 2a$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.

Tʜᴜs,

$\rm :\longmapsto\:\dfrac{a_6}{a_{21}}$

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

$\rm \:  =  \:\dfrac{2a + 10d}{2a + 40d}$

$\rm \:  =  \:\dfrac{d + 10d}{d + 40d}$

$\rm \:  =  \:\dfrac{11d}{41d}$

$\rm \:  =  \:\dfrac{11}{41}$

• Hence, Option (d) is correct.