The sum of $49$ consecutive integers is $7^{5}$ . What is their median ?

7

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

7^{2}

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

7^{3}

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

7^{4}

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

Open in App

Solution

The correct option is C $7^{3}$
C) let these consecutive numbers are $X - 24, X - 23, . . ., X, X + 1, . . ., X + 24$
here X is median
then sum of these numbers is $49 X =$ $7^{5}$
$X =$ $7^{3}$
since Xis median thus $X =$ $7^{3}$

Suggest Corrections

Similar questions

\frac{1}{7} + \frac{2}{7^{2}} + \frac{1}{7^{3}} + \frac{2}{7^{4}} + . . .

\frac{4}{7} - \frac{5}{7^{2}} + \frac{4}{7^{3}} - \frac{5}{7^{4}} + \frac{4}{7^{5}} - \frac{5}{7^{6}} + . . . .

\frac{1}{7} + \frac{2}{7^{2}} + \frac{1}{7^{3}} + \frac{2}{7^{4}} + . . . .

1 + \frac{4}{7} + \frac{9}{7^{2}} + \frac{16}{7^{3}} + \frac{25}{7^{4}} + . . . . . .

1 + \frac{4}{7} + \frac{9}{7^{2}} + \frac{16}{7^{3}} + \frac{25}{7^{4}} + \dots

Join BYJU'S Learning Program