Test: Geometric Progressions - JEE MCQ

ar⁶/ar² = 16
r⁴ = 16
r⁴ = (2)⁴
r = 2
S₉ = a(rⁿ - 1)
2555 = a((2)⁹ - 1)
2555 = a(512-1)
2555/511 = a
a = 5

The geometric mean of two numbers, say x, and y is the square root of their product x * y.
 Mean = [x * y]^1/2
= [5 * 8]^1/2
= [40]^1/2
= 2(10)^1/2

ar⁹ = 1/(2)⁸ -----------------------(1)
ar³ = 1/(2)² ------------------------------(2)
Comparing eq (1) & (2)
1/(2)⁸ r⁹ = 1/(2)^2 r⁴
1/r⁵ = (2)⁵
r⁵ = 1/(2)⁵
r = 1/2

The expression depicts the ratio of two numbers, if the ratio between the numbers is constant, then it will definately form a GP.

g_n = 6( 3^n-1) it is a geometric expression with coefficient of constant as 3^n-1.So it is GP with common ratio 3.

8(1) + 8(11) + 8(111) + 8(1111) +........up to n times
= 8{ 1 + 11 + 111 + 1111 + ....... up to n times}
= 8/9{9 + 99 + 999 + 9999 + ..... up to n times}
= 8/9 {(10-1 )+ (100-1) + (1000-1) + (10000-1) + ........up to n times}
= 8/9 {(10 + 100 + 1000 + 10000 +...... n times) - (n×1)}
= 8/9{( (10×(10ⁿ -1))/(10-1)) -n}
= 80/81(10ⁿ - 1) - n
=  8/81(10ⁿ⁺¹ - 10 -9n)

Given sequence : 2, 2√2, 4….
First term a₁ = a = 2 and 2nd term a₂ = 2√2, then
Common ratio r = a₂/a = (2√2)/2
Let a_n = 64
∴ ar^(n-1) = 64
⇒ 2.(√2)^(n-1) = 32
⇒ (2)^(n-1)/2 = 32
∴ (2)^(n-1)/2 = (2)⁵
⇒ (n − 1)/2 = 5,
⇒ n = 11

Sum (S_n) = a x (rⁿ -1)/(r-1)
5460 = 4 x (4ⁿ -1)/3
16380 = 4ⁿ⁺¹ - 4

16384 = 4ⁿ⁺¹

4⁷ = 4ⁿ⁺¹

7 = n + 1

n = 6

an = ar^(n-1)
 Given, a = √3, r = 3/√3
​r = √3
a₅₀ = ar^(n-1)
= (√3)(√3)^(50-1)
= (√3)(√3)⁴⁹
= (√3)⁵⁰