Question

Let $a_1, a_2, a_3, ...$ be terms of an A.P.. If $\frac{a_1+a_2+…+a_p}{a_1+a_2+…+a_q}=\frac{p^2}{q^2}$, where $p \neq q$, then $\frac{a_6}{a_{21}}$ is equal to

• A. 11/41
• B. 21/6
• C. 6/21
• D. 41/11

Answer 1

Answer: (A)

11/41

Answer 2

Let $a$ be the first term and $d$ be the common difference of the Arithmetic Progression (A.P.). The sum of the first $n$ terms of an A.P. is given by the formula: $S_n = \frac{n}{2}[2a + (n-1)d]$ We are given the relation: $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^2}{q^2}$ Substituting the sum formula, we get: $\frac{\frac{p}{2}[2a + (p-1)d]}{\frac{q}{2}[2a + (q-1)d]} = \frac{p^2}{q^2}$ $\frac{p[2a + (p-1)d]}{q[2a + (q-1)d]} = \frac{p^2}{q^2}$ Since $p \neq q$, we can assume $p \neq 0$ and $q \neq 0$. We can simplify the equation by cancelling $p$ from the numerator and $p^2$ from the right side, and $q$ from the denominator and $q^2$ from the right side: $\frac{2a + (p-1)d}{2a + (q-1)d} = \frac{p}{q}$ Cross-multiplying gives: $q[2a + (p-1)d] = p[2a + (q-1)d]$ $2aq + q(p-1)d = 2ap + p(q-1)d$ $2aq + pqd - qd = 2ap + pqd - pd$ Subtracting $pqd$ from both sides: $2aq - qd = 2ap - pd$ Rearranging the terms to group $a$ and $d$: $2aq - 2ap = qd - pd$ $2a(q - p) = d(q - p)$ Since $p \neq q$, we know that $q - p \neq 0$. Therefore, we can divide both sides by $(q - p)$: $2a = d$ This establishes a relationship between the first term and the common difference.

Now we need to find the ratio $\frac{a_6}{a_{21}}$. The $n$-th term of an A.P. is given by $a_n = a + (n-1)d$. So, the 6th term is $a_6 = a + (6-1)d = a + 5d$. And the 21st term is $a_{21} = a + (21-1)d = a + 20d$.

Substitute the relation $d = 2a$ into the expressions for $a_6$ and $a_{21}$: $a_6 = a + 5(2a) = a + 10a = 11a$ $a_{21} = a + 20(2a) = a + 40a = 41a$ Now, we can find the ratio: $\frac{a_6}{a_{21}} = \frac{11a}{41a}$ Assuming $a \neq 0$ (otherwise all terms would be zero, making the original ratio undefined), we can cancel $a$: $\frac{a_6}{a_{21}} = \frac{11}{41}$

Alternatively, we can use the property that if $\frac{S_p}{S_q} = \frac{p^2}{q^2}$ for an A.P., then $\frac{a_m}{a_n} = \frac{2m-1}{2n-1}$. In this case, we need $\frac{a_6}{a_{21}}$, so $m=6$ and $n=21$. $\frac{a_6}{a_{21}} = \frac{2(6)-1}{2(21)-1} = \frac{12-1}{42-1} = \frac{11}{41}$