If in an A.P. \(\frac{a_7}{a_{10}}\) = \(\frac{5}{7},\) find \(\frac{

1.	If in an A.P. \(\frac{a_7}{a_{10}}\) = \(\frac{5}{7},\) find \(\frac{a_4}{a_7}\)
Answer» Given, \(\frac{a_7}{a_{10}}\) = \(\frac{5}{7}\) Let the first term and common difference of AP be ′A’ and ′D’, respectively. ∴ \(\frac{A\, +\, 6D}{A\, +\, 9D}\) = \(\frac{5}{7}\) ⇒ 7A + 42D = 5A + 45D ⇒ 7A = 3D ⇒ A \(\frac{3}{2}\) D ….. (i) Now, \(\frac{a_4}{a_7}\) = \(\frac{A\,+\,3D}{A\,+\,6D}\) = \(\frac{\frac{3}{2}D}{\frac{3}{2}D}\) = \(\frac{3D}{6D}\) … . . (using(i)) = \(\frac{9D}{15D}\) = \(\frac{3}{5}\)

If in an A.P. \(\frac{a_7}{a_{10}}\) = \(\frac{5}{7},\) find \(\frac{a_4}{a_7}\)

Answer»

Given, \(\frac{a_7}{a_{10}}\) = \(\frac{5}{7}\)

Let the first term and common difference of AP be ′A’ and ′D’, respectively.

∴ \(\frac{A\, +\, 6D}{A\, +\, 9D}\) = \(\frac{5}{7}\)

⇒ 7A + 42D = 5A + 45D

⇒ 7A = 3D

⇒ A \(\frac{3}{2}\) D ….. (i)

Now, \(\frac{a_4}{a_7}\) = \(\frac{A\,+\,3D}{A\,+\,6D}\)

= \(\frac{\frac{3}{2}D}{\frac{3}{2}D}\) = \(\frac{3D}{6D}\) … . . (using(i))

= \(\frac{9D}{15D}\)

= \(\frac{3}{5}\)