Question

Let a1, a2, a3, ..... be the terms of an ap . if a1+a2+a3+....ap/a1+a2+a3+.....aq = p^2/q^2 where pnot equal to q , find the value of a6/a21

Answer 1

The sum of $n$ terms of an AP with first term $a$ and common difference $d$ is

$S_n=\frac{n[2a+(n-1)d]}{2}$.

So,

$S_p=\frac{p[2a+(p-1)d]}{2}\\ S_q=\frac{q[2a+(q-1)d]}{2}$

It is given that

$\frac{S_p}{S_q}=\frac{p^2}{q^2}\\ \frac{\frac{p[2a+(p-1)d]}{2}}{\frac{q[2a+(q-1)d]}{2} }=\frac{p^2}{q^2}\\ \frac{2a+(p-1)d}{2a+(q-1)d }=\frac{p}{q}\\ [2a+(p-1)d]q=[2a+(q-1)d]p\\ 2a(q-p)=d(q-p)\\ 2a=d$

Now,

$\frac{a_6}{a_{21}} =\frac{a+5d}{a+20d} \\ \frac{a_6}{a_{21}} =\frac{a+5(2a)}{a+20(2a)} \\ \frac{a_6}{a_{21}} =\frac{11a)}{41a} \\ \frac{a_6}{a_{21}} =\frac{11}{41}$

Answer 2

Answer:

Value of $\frac{a_6}{a_{21}}\:is\:\frac{11}{41}$.

Step-by-step explanation:

Given:

$\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2}$

we know that,

$S_n=\frac{n}{2}(2a+(n-1)d)$

So,

$\frac{\frac{p}{2}(2a+(p-1)d)}{\frac{q}{2}(2a+(q-1)d)}=\frac{p^2}{q^2}$

$\frac{2a+(p-1)d}{2a+(q-1)d}=\frac{p}{q}$

$\frac{2a+pd-d}{2a+qd-d}=\frac{p}{q}$

$(2a+pd-d)q=(2a+qd-d)p$

$2aq+pqd-qd=2ap+pqd-pd$

$2aq-2ap=-pqd+qd+pqd-pd$

$2a(q-p)=(q-p)d$

$2a=d$

Now, using $a_n=a+(n-1)d$

$\frac{a_6}{a_{21}}=\frac{a+(6-1)d}{a+(21-1)d}$

$\frac{a_6}{a_{21}}=\frac{a+5(2a)}{a+20(2a)}$

$\frac{a_6}{a_{21}}=\frac{a+10a}{a+40a}$

$\frac{a_6}{a_{21}}=\frac{11a}{41a}$

$\frac{a_6}{a_{21}}=\frac{11}{41}$

Therefore, Value of $\frac{a_6}{a_{21}}\:is\:\frac{11}{41}$