Progression (Sequence and Series)Hard
Question
Let S1, S2...... be squares such that for each n ≥ the length of a side of Sn equals the lenght of a disgonal of Sn+1. If the length of a side of S1 is 10 cm, then for which of the following values of n is the area of Sn less than 1 sq. cm ?
Options
A.7
B.8
C.9
D.10
Solution
Let n a denotes the length of side of the square Sn
We are given n a = length of diagonal of n S1+n.
⇒ an = √2 an+1
⇒ an+1 =
This shows that a1, a2, a3 ..... form a GP with common ratio 1/√2.
Therefore, an = a1
⇒ an = 10
(∵ a1 = 10 given)
⇒ an2 = 100
⇒
≤ 1 (∵ an2 ≤ 1 given)
⇒ 100 ≤ 2n-1
This is possible for n ≥ &
So, (b),(c),(d) are the answers.
We are given n a = length of diagonal of n S1+n.
⇒ an = √2 an+1
⇒ an+1 =
This shows that a1, a2, a3 ..... form a GP with common ratio 1/√2.
Therefore, an = a1
⇒ an = 10
(∵ a1 = 10 given)⇒ an2 = 100

⇒
≤ 1 (∵ an2 ≤ 1 given)⇒ 100 ≤ 2n-1
This is possible for n ≥ &
So, (b),(c),(d) are the answers.
Create a free account to view solution
View Solution FreeMore Progression (Sequence and Series) Questions
Every term of an infinite GP is thrice the sum of all the successive terms. If the sum of first two terms is 15, then th...A G.P. consists of 2n terms. If the sum of the terms occupying the odd places is S1 and that of the terms at the even pl...If sum of infinite G.P. is x and sum of square of its terms is y, then common ratio is-...With usual notations in ᐃABC, it is given that r1, r2, r3 are in harmonic progression. The perimeter of triangle i...The sum of the first 100 terms common to the series 17, 21, 25, ....... and 16, 21, 26, ........is -...