First slide

Arithmetic progression

Question

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

Moderate

A

100
B

99
C

102
D

101

Solution

$\begin{array}{l} \underline{{(\frac{8}{5})}^{2} + {(\frac{12}{5})}^{2} + {(\frac{16}{5})}^{2} + . . + {(\frac{44}{5})}^{2}} (\sin c e 10' t h t e r m o f t h e A . P = \frac{8 + 9 \cdot 4}{5}) \\ = \frac{16}{25} (2^{2} + 3^{2} + . . + 11^{2}) \\ = \frac{16}{25} [\frac{11 (11 + 1) (22 + 1)}{6} - 1] (\sin c e s u m o f t h e s q u a r e s o f f i r s t n n a t u r a l n u m b e r s = \frac{n (n + 1) (2 n + 1)}{6}) \\ = \frac{16}{25} \times 505 = \frac{16}{5} \times 101 = \frac{16}{5} m \\ \Rightarrow m = 101 \end{array}$

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