Question

Let a₁, a₂,a_3, …………be terms of an A.P. If

$\frac{a_{1} + a_{2} + .... + a_{p}}{a_{1} + a_{2} + .... + a_{q}}$= $\frac{p^{2}}{q^{2}}$, p ¹ q, then $\frac{a_{6}}{a_{21}}$ equals –

• A. $\frac{41}{11}$
• B. $\frac{7}{2}$
• C. $\frac{2}{7}$
• D. $\frac{11}{41}$

Answer 1

Answer: (D)

$\frac{11}{41}$

Answer 2

$\frac{a_{1} + a_{2} + .... + a_{p}}{a_{1} + a_{2} + ..... + a_{q}}$= $\frac{p^{2}}{q^{2}}$

$\frac{a_{1} + a_{2} + .... + a_{6}}{a_{1} + a_{2} + ..... + a_{q}}$= $\frac{6^{2}}{q^{2}}$ …(i)

$\frac{a_{1} + a_{2} + .... + a_{5}}{a_{1} + a_{2} + ..... + a_{q}}$= $\frac{5^{2}}{q^{2}}$ …(ii)

on subtracting eqⁿ. (i) – (ii)

$\frac{a_{6}}{a_{1} + .... + a_{q}}$= $\frac{6^{2}–5^{2}}{q^{2}}$ ….(iii)

Similarly$\frac{a_{21}}{a_{1} + .... + a_{q}}$= $\frac{(21)^{2}–(20)^{2}}{q^{2}}$....(iv)

Divide (iii) and (iv) $\frac{a_{6}}{a_{21}}$ = $\frac{11}{41}$