If $S_{A} = 2 m + (2 m + 1) + (2 m + 2) + . . . + 4 m$ and $S_{B} = (2 m + 1) + (2 m + 3) + (2 m + 5) + . . . + (4 m - 1),$ then

\frac{S_{A}}{S_{B}} = 2 + \frac{1}{m}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{S_{A}}{S_{B}} = \frac{4}{3} + \frac{2}{3 m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{S_{A}}{S_{B}} = \frac{4}{3} + \frac{1}{m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{S_{A}}{S_{B}} = \frac{1}{3} + \frac{1}{m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{S_{A}}{S_{B}} = 2 + \frac{1}{m}$
$S_{A} = 2 m + (2 m + 1) + (2 m + 2) + . . . + 4 m$
Number of terms in the series $S_{A}$ is $2 m + 1$
$S_{A} = \frac{2 m + 1}{2} (4 m + 2 m) ∴ S_{A} = 3 m (2 m + 1)$

$S_{B} = (2 m + 1) + (2 m + 3) + . . . + (4 m - 1)$
Number of terms in the series $S_{B}$ is $m$
$S_{B} = \frac{m}{2} [2 m + 1 + 4 m - 1] ∴ S_{B} = 3 m^{2}$

Therefore,
$\frac{S_{A}}{S_{B}} = \frac{2 m + 1}{m} = 2 + \frac{1}{m}$

Suggest Corrections

Similar questions

S_{A} = 2 m + (2 m + 1) + (2 m + 2) + . . . + 4 m

S_{B} = (2 m + 1) + (2 m + 3) + (2 m + 5) + . . . + (4 m - 1),

2 m / s

A body has an initial velocity of $4 m s^{- 1}$ and deceleration of $4 m s^{- 2}$ , then what is its final velocity after travelling 2 m?

A body has an initial velocity of 4 m/s and deceleration of 4 $m / s^{2}$ , then what is its final velocity after travelling 2m?

\frac{4 m + 4}{2 m + 1} - \frac{3}{4}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program