Let $a_1,a_2,a_3....$ be an arithmetic progression.

Question

Question is

Let $a_1,a_2,a_3....$ be an arithmetic progression. If $\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2},p\not=q$, then find $\frac{a_6}{a_{21}}$

Answer: $\frac{11}{41}$

All options: $\frac{7}{2}$, $\frac{2}{7}$, $\frac{11}{41}$, $\frac{41}{11}$

My approach to the question is

Let $S_r$ be sum of first $r$ terms of this AP

Method 1

we know $\frac{S_p}{S_q}=\frac{p^2}{q^2}$

So,

$\frac{\frac{p}{2}(a_1+a_p)}{\frac{q}{2}(a_1+a_q)}=\frac{p^2}{q^2}$

$\frac{a_1+a_p}{a_1+a_q}=\frac{p}{q}$

After that I am stuck.

Method 2

But after taking another look at the question, I observed

$\frac{S_6}{S_{21}}=\frac{6^2}{21^2}=\frac{36}{441}$

Also,

$\frac{S_5}{S_{20}}=\frac{5^2}{20^2}=\frac{25}{400}$

And we know $S_r-S_{r-1}=a_r$

Then subtracting numerators and denominators of both

$\frac{S_6-S_5}{S_{21}-S_{20}}=\frac{a_6}{a_{21}}=\frac{36-25}{441-400}=\frac{11}{41}$

which we can't always do in maths, I cannot understand why Method 2 gave correct answer.

Answer 1

You can find from your first method!

$a_p = a_1+(p-1)d \\a_q = a_1+(q-1)d$

$$\implies\frac{2a_1+(p-1)d}{2a_1+(q-1)d}=\frac{p}{q}\implies2a_1q+pdq-dq=2a_1p+pdq-dp$$

$$\implies(2a-d)(q-p)=0 \iff q=p \text{ or } d=2a$$

As, we consider different $p$ and $q$, we have

$$\frac{a_6}{a_{21}} = \frac{a_1+5d}{a_1+20d}=\frac{11a_1}{41a_1}=\frac{11}{41}$$

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Your method $2$ works for this case as long as you don't simplify the fractions. The reason behind it is,

$$\frac{s_p - s_{p-1}}{s_1-s_{q-1}}=\frac{a_p}{a_q}=\frac{a+(p-1)2a}{a+(q-1)2a}=\frac{2p-1}{2q-1}\equiv\frac{p^2-(p-1)^2}{q^2-(q-1)^2}$$ So don't use the second method as it may not work for other questions.

Answer 2

We have that $$\frac{a_1+a_2}{a_1}=\frac{2^2}{1^2}=4$$ $$a_1+a_2=4a_1$$ $$a_2=3a_1$$ So the common difference is $$d=a_2-a_1=2a_1$$ Hence our final answer is $$\frac{a_6}{a_{21}}=\frac{a_1+(6-1)(2a_1)}{a_1+(21-1)(2a_1)}=\frac{11}{41}$$

Answer 3

I also got $\frac{11}{41}.$

We have: $$\frac{(2a_1+(p-1)d)p}{(2a_1+(q-1)d)q}=\frac{p^2}{q^2},$$ which gives $$d=2a_1$$ and $$\frac{a_6}{a_{21}}=\frac{a_1+5\cdot2a_1}{a_1+20\cdot2a_1}=\frac{11}{41}.$$

Answer 4

$$\dfrac{p^2}{q^2}=\dfrac{p(2a_1+(p-1)d}{q(2a_1+(q-1)d}$$

$$\implies \dfrac pq=\dfrac{a_1+d(p-1)/2}{a_1+d(q-1)/2}$$

Set $\dfrac{p-1}2=6-1$

and $\dfrac{q-1}2=21-1$