Sequences and Series 5 Question 19

10. Let $S _1, S _2, \ldots$ be squares such that for each $n \geq 1$ the length of a side of $S _n$ equals the length of a diagonal of $S _{n+1}$. If the length of a side of $S _1$ is $10 cm$, then for which of the following values of $n$ is the area of $S _n$ less than $1 sq cm$ ?

(1999, 3M)

(a) 7

(b) 8

(c) 9

(d) 10

Analytical & Descriptive Questions

Show Answer

Solution:

  1. Let $S _{10}$ be the sum of first ten terms of the series. Then, we have

$S _{10}=1 \frac{3}{5}^{2}+2 \frac{2}{5}^{2}+3 \frac{1}{5}^{2}+4^{2}+4 \frac{4}{5}^{2}$

$+\ldots$ to 10 terms

$$ \begin{aligned} & =\frac{8}{5}^{2}+\frac{12}{5}^{2}+\frac{16}{5}^{2}+4^{2}+\frac{24}{5}^{2}+\ldots \text { to } 10 \text { terms } \\ & =\frac{1}{5^{2}}\left(8^{2}+12^{2}+16^{2}+20^{2}+24^{2}+\ldots \text { to } 10 \text { terms }\right) \\ & =\frac{4^{2}}{5^{2}}\left(2^{2}+3^{2}+4^{2}+5^{2}+\ldots \text { to } 10 \text { terms }\right) \\ & =\frac{4^{2}}{5^{2}}\left(2^{2}+3^{2}+4^{2}+5^{2}+\ldots+11^{2}\right) \\ & =\frac{16}{25}\left(\left(1^{2}+2^{2}+\ldots+11^{2}\right)-1^{2}\right) \\ & =\frac{16}{25} \frac{11 \cdot(11+1)(2 \cdot 11+1)}{6}-1 \\ & =\frac{16}{25}(506-1)=\frac{16}{25} \times 505 \Rightarrow \frac{16}{5} m=\frac{16}{25} \times 505=101 \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane