Sequences And Series- 2 - Free MCQ Practice Test with solutions, JEE Chapter

a₁ + a_2n = a₂ + a_2n-1 = a_n + a_n-1 = K(say)

a + f = b + e = c + d

Let
If α, β,and γ are the three roots then we must have
ax³ + bx² + cx + d = a(x−α)(x−β)(x−γ)
Comparing coefficients of various power of x on both sides leads to
α + β + γ = −b/a −−−−(1)
αβ + βγ + γα = c/a −−−−(2)
αβγ= −d/a−−−(3)
On substituting β = αr and γ = αr2 we get
α(1+r+r²) = −b/a−−−(4)
α²(r+r² + r³) = c/a−−(5)
α³/r³ = -d/a−−−−−(6)
Now, dividing (5) by (4) to αr=− c/b and substituting this in(6)weget
(−c/b)³ = -d/a
⇒ c³/b³ = -d/a
∴ c³a = b³d

Let x and y be two positive numbers. Since a is the single A.M. between x and y
∴ 
It is given that p and q are two geometric means between x and y common ratio r given by 
r = (y/x)^1/3
∴ p = xr = x^2/3 y^1/3 and q = xr² = x^1/3y^2/3
⇒ p³ +q³ = x²y + xy² = xy(x+y) = 2axy = 2a pq

6, a bare in H.P 
 are in A.P.

(2+0) + (2+1²) + (2+2²) + (2 3²)+...
∴ t_n = 2+(n-1)²

a + b = 16 and ab = 16 and 2/x = 1/a + 1/b

a,b,c are in H.P
∴1/a , 1/b , 1/care in A.P
∴2/b= 1/a+1/c_______ (1)
∴ x/a+y/b+1/c=0 ________ (2)
∴ From (1)
1/a+1/c−2/b=0 ______ (3)
∴ From (2) & (3)
We infer that,
(x=1,y=−2)
∴ Point is (1,−2)

Given that  and a, y, z, b are in G.P. 

a + b = a₁ +a_2n = a₂ + a_2n-1 = and ab = g₁g_2n = g₂g_2n-1 .... and

cos(x-y), cosx, cos(x+y)
cos x = [2cos(x-y)cos(x+y)]/2cosx cosy
= [cos²x cos²y - sin²x sin²y]/cosx cosy
=> cos²x cosy = cos²x cos²y - sin²x sin²y
= cos²x cos²y(1 - cos²x - cos²y + cos²x  cos²y)
= cos²x cosy = cos²x + cos²y - 1
cos²x = cos(y+1)
cos²x = 2cos²(y/2)
cosx sec(y/2) = +- (2)^1/2

b² = ac 
ax² + 2bx+c = 0 

2a/2be^y = 2b/2ce^y = 2c/2de^y
= a/be^y = b/ce^y = c/de^y = k
a/b = b/c = c/d = ke^y = k'
a=bk', b=ck',  c = dk'
b = (dk')k'
b = d(k')²
a = bk' = d(k')² k'
= d(k')³
a,b,c,d ⇒ d(k')³, d(k')², d(k'), d
G.P. sequence

Roots are real and equal
⇒ (a² + b² + c²) (b² + c² + d²)-(ab + bc + cd)² = 0 
⇒ b² = ac, c² = bd, ac = bd
⇒ a, b, c, d are in G.P

1
2 3
4 5 6
7 8 9 10
..........
Series for starting number:
1, 2, 4, 7, ...
Let  Tn = an² + bn + c
a+b+c=1                         ......(1)
4a+2b+c=2                   ......(2)
9a+3b+c=4                   .......(3)
From eq (1), (2), (3), we get:
a= 1/2, b= −1/2, c=1
∴ T_n = n²/2 − n/2 + 1
∴ T₅₀ = (50)²/2 − 50/2 + 1 = 1226
Now, elements in 50th group:
1226, 1227, 1228,.....(50 terms)
∴ Sn = 50/2[2×1226+(50−1)×1]
= 62525

a + ar + ar² = x.ar
Or, r² + r (1 - x) + 1 = 0, r is real
Δ > 0 i.e. (1 - x )² - 4 > 0
Or, x² - 2x - 3 > 0
Or, (x + 1) ( x - 3) > 0
⇒ x < -1 or x > 3

2b = a + c ….. (i)
b⁴ = a² c² or b² = ± ac .....(ii)
Here 
a+b+c = 3/2
⇒ 3b = 3/2 or b = 1/2
from (i)
From (i), a+c = 1

From (ii) 
Solving this we get

From synopsis

For the first AP, a =2, d = 5
Hence, any nth term would be given by 2+(n-1)5 = 5n-3
Also, since 500 is the last term, so, 500 = 5n-3 i.e. maximum value of n can be 100
 
For the second AP, a = 1, d = 7
Hence any mth term would be given by 1+(m-1)7 = 7m - 6
Also, since 300 is the last terms, so, 300 = 7m-6 i.e. maximum value of m can be 41
 
To find the terms common to both the APs, we can equate the nth term of first AP to the mth term of the second AP
5n -3 = 7m-6
5n = 7m-3
 
Now, since n is a whole number, so 7m-3 needs to be divisible by 5
So, then, m can be equal to all multiples of 5 till 102 i.e. 5, 10, 15, 20, 25, 30, ......100
So, the number of terms common to the 2 APs would be 60.

t is given that
log2,log(2^x -l),log (2^x + 3) are in A.P.
⇒ 2log (2^x-l) = log 2 + log (2^x + 3)
⇒(2^x -1)² = 2 (2^x + 3)
⇒ (2^x)² -4(2^x)-5 = 0
⇒ (2^x - 5)(2^x + 1) = 0 ⇒ 2^x = 5 ⇒ x = log₂5

 1/a_r = 1/a_1-r + d
d a_r a_r-1 = a_r-1 - a_r
ar a_r-1 = (a_r-1 - ar)/d
Σ(r = 2 to n) ar ar-1 = (a₁ - a_n)/d
⇒ 1/a_n = 1/a₁ + (n-1)d
⇒ (a₁ - a_n)/d = (n-1)a₁a_n

Let ‘d’ is common difference of A.P
∴ 3 = a₁₀
 
Let ‘D’ is common difference of 
∴ 

Sum of Squares:  n(n+1)(2n+1)6 
Therefore the denominator can be simplified using this formula,
= 3 + (5⋅6)/(2⋅3⋅5) + (7⋅6)/(3⋅4⋅7) + ⋯+ (101⋅6)/(50⋅51⋅101) 
= 3 + 6[(2⋅3)+6(3⋅4)+⋯+6/(50⋅51)] 
= 3 + 6[1/(2⋅3)+1/(3⋅4)+⋯+1/(50⋅51)] 
= 3 + 6[1/2−1/3+1/3+1/4+⋯+1/50−1/51] 
= 3 + 6[1/2−1/51] 
= 3 + 6(51−2)/(51⋅2) 
= 3 + 49/17 
= 100/17

= r(-a)^r + a(r+1) (-a)r
= r(-a)^r - (r+1) (-a)^(r+1)
T_r = V_r - V_r+1
T₁ = V₁ - V₂
T₂ = V₂ - V₃
T₃ = V₃ - V₄
.
.
.
.
.
T_n = V_n - V_(n+1)
S = (r = 1 to n) T_r = V₀ - V_(n+1)
= (-1)ⁿ (n+1)(-a)⁽ⁿ⁺¹⁾
 

9/4 = 1 + 2r + 3r² + 4r³ ⋯→(1)
By mutiplying above equation by r, we get,
9/4r = r + 2r² + 3r³ + 4r⁴ ⋯→(2)
Subtracting (2) from (1) by shifting one place, we get
9/4(1−r) = 1 + r + r² + r³ + r⁴ .......
The above equation becomes ,
9/4(1−r) = 1/(1−r)
​(1−r)² = 4/9
Hence, we get
r = ⅓