# Progression & Series

Algebra

## Which number is divisible by all numbers from 1 to 10 ?

Q: Which number is divisible by all numbers from 1 to 10 ?

Ans: If we multiply two numbers then its product will be divisible by both the numbers . e.g 5 × 7 = 35 .

Similarly , If we multiply all the numbers from 1 to 10 then its product will be divisible by all the numbers 1 to 10

i.e. 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 .

But number 8 is divisible 4 and 4 is divisible by 2 , similarly 10 is divisible by 5 so by considering these concepts the product of 5 × 7 × 8 × 9 will be divisible by all the numbers from 1 to 10 .

Hence the Required number = 5 × 7 × 8 × 9 = 2520

## A geometric progression of real numbers is such that the sum of its first four terms is equal to 30 and the sum of the squares of the first four terms is 340. Then

Q: A geometric progression of real numbers is such that the sum of its first four terms is equal to 30 and the sum of the squares of the first four terms is 340. Then

(A) two such G.P. are possible

(B) it must be a decreasing G.P.

(C) the common ratio is always rational

(D) the first term is always an even integer

Ans: (A), (C), (D)

Let the G.P be a, ar, ar2, ….

We have a + ar + ar2 + ar3 = 30

a2 + a2r2 + a2r4 + a2r6 = 340

$\large \frac{(1 + r + r^2 + r^3)^2}{1 + r^2 + r^4 + r^6} = \frac{900}{340} = \frac{45}{17}$

$\large \frac{(r^4 -1)^2 (r^2 -1)}{(r-1)^2 (r^8 -1)} = \frac{45}{17}$  ; (since ,  r ≠ 1)

$\large \frac{(r^4 -1) (r+1)}{(r-1) (r^4 +1)} = \frac{45}{17}$

$\large \frac{r^4 + 2r^3 + 2r^2 + 2r + 1}{r^4 +1} = \frac{45}{17}$

14 r4 – 17 r3 – 17 r2 – 17 r + 14 = 0

Dividing by r2 , we get 14 – 17 – 17 = 0

$\large 14 (r^2 + \frac{1}{r^2}) – 17(r + \frac{1}{r}) -17 = 0$

$\large 14 [(r + \frac{1}{r})^2 -2 ] – 17(r + \frac{1}{r}) -17 = 0$

Putting , $r + \frac{1}{r} = y$

14y2 – 17 y – 45 = 0

$\large y = \frac{17 \pm 53}{28}$

=> y = 5/2 , – 9/7

=> r + 1/r = 5/2 => r = 2, 1/2 and r + 1/r = –9/7 has no real roots.

Putting r = 2 in (1), we get a = 2 and putting r = 1/2 in (1), we get a = 16

Then the progression is 2, 4, 8, 16, 32, 64, …….. or, 16, 8, 4, 2, 1, 1/2 , ….

## If $a^2 + b^2 = 1$ , $m^2 + n^2 = 1$ , then

Q: If $\large a^2 + b^2 = 1$ , $\large m^2 + n^2 = 1$ , then

(A) |am + bn| ≤ 1

(B) |am – bn| ≥ 1

(C) |am + bn| ≥  1

(D) none of these

Sol. (A). If a2 + b2 = 1, then obviously we can find an angle a such that

a = cos α , b = sin α

but we can have β , such that m = cos β , n = sin β

∴ |am + bn| = |cos α cos β + sin α sin β| = |cos (α – β)| ≤ 1

## Let p , q , r ∈ R+ and 27 pqr  ≥ (p + q + r)^3 and 3p + 4q + 5r = 12 then p^3 + q^4 + r^5 is equal to

Q: Let p , q , r ∈ R+ and 27 pqr  ≥ (p + q + r)3 and 3p + 4q + 5r = 12 then
p3 + q4 + r5 is equal to

(A) 3

(B) 6

(C) 2

(D) none of these

Sol. (A). 27 pqr  ≥ (p + q + r)3

$\large (pqr)^{1/3} \ge \frac{p+q+r}{3}$

⇒  p = q = r

Also 3 p + 4q + 5r =12

⇒ p = q = r = 1

## The greatest value of x^2 y^3 is, where x > 0 and y > 0 are connected by the relation 3x + 4y = 5

Q: The greatest value of x2y3 is, where x > 0 and y > 0 are connected by the relation 3x + 4y = 5

(A) 3/16

(B) 1/16

(C) 3/14

(D) none of these

Sol: (A). We are given 3x + 4y = 5 …(1)

Now, product x2y3 will be maximum when $(\frac{3}{2}x)^2 (\frac{4}{3}y)^3$ is maximum. Above is product of 5 factors whose sum is

$\large 2.\frac{3}{2}x + 3.\frac{4}{3}y = 3x + 4y = 5 = constant$

Hence the product will be maximum when all the factors are equal

i.e. $\frac{3x}{2} = \frac{4 y}{3} = \frac{3x+4y}{2+3} = \frac{5}{5} = 1$

⇒ x = 2/3 , y = 3/4

∴ Maximum value of x2y3 $\large = (\frac{2}{3})^2 (\frac{3}{4})^3 = \frac{3}{16}$