If the sum of the first ten terms of the series 13-5 2 - 22-5 2 - 31-5 2 - 42 - 44-5 2-is 16-5m- then m is equal to-

( 8/5 )^2 + ( 12/5 )^2 + ( 16/5 )^2 + ( 20/5 )^2 ...+ ( 44/5 )^2 S = 16/25(2^2 + 3^2 + 4^2 + ...+11^2) =16/25 ( 11(11+1)(22+1)/6 1 ) = 16/25× 505 = 16/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

Sigma n2

If the sum of...

Question

If the sum of the first ten terms of the series ${(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + . . . . .,$ is $\frac{16}{5} m,$ then m is equal to:

100

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

102

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

101

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

Open in App

Solution

The correct option is D 101
${(\frac{8}{5})}^{2} + {(\frac{12}{5})}^{2} + {(\frac{16}{5})}^{2} + {(\frac{20}{5})}^{2} . . . + {(\frac{44}{5})}^{2} S = \frac{16}{25} (2^{2} + 3^{2} + 4^{2} + . . . + 11^{2}) = \frac{16}{25} (\frac{11 (11 + 1) (22 + 1)}{6} - 1) = \frac{16}{25} \times 505 = \frac{16}{5} \times 101 \Rightarrow \frac{16}{5} m = \frac{16}{5} \times 101 \Rightarrow m = 101.$

Suggest Corrections

Similar questions

Q. If the sum of the first ten terms of the series

{(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + . . . . . . ., i s \frac{16}{5} m,

then m is equal to:

Q. If the sum of the first ten terms of the series

{(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + \dots \dots i s \frac{16}{5} m

, then m is equal to

Q. If the sum of the first ten terms of the series

{(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + . . . . .,

\frac{16}{5} m,

then m is equal to:

Q. If the sum of the first ten terms of the series

{(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + \dots \dots i s \frac{16}{5} m

, then m is equal to

Q. If the sum of the first ten term of the series

{(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + . . . .,

\frac{16 m}{5}

, then m is equal to

Join BYJU'S Learning Program

If the sum of the first ten terms of the series (135)2+(225)2+(315)2+42+(445)2+.....,is 165m, then m is equal to:

The correct option is D 101(85)2+(125)2+(165)2+(205)2...+(445)2S=1625(22+32+42+...+112)=1625(11(11+1)(22+1)6−1)=1625×505=165×101⇒165m=165×101⇒m=101.

If the sum of the first ten terms of the series ${(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + . . . . .,$ is $\frac{16}{5} m,$ then m is equal to: