DXC Technology Series and Progression Quiz-1

Here, a = 11, d = 17 – 11 = 23 – 17 = 29 – 23 = 6
We know that nth term of an AP is a + (n – 1) d
=> nth term for the given AP = 11 + (n – 1) 6
=> nth term for the given AP = 5 + 6 n
We can verify the answer by putting values of ‘n’.
=> n = 1 -> First term = 5 + 6 = 11
=> n = 2 -> Second term = 5 + 12 = 17
=> n = 3 -> Third term = 5 + 18 = 23
and so on …

From the above question,
=> nth term for the given AP = 5 + 6 n
=> First term = 5 + 6 = 11
=> Tenth term = 5 + 60 = 65
=> Sum of 10 terms of the AP = 0.5 n (first term + last term) = 0.5 x 10 (11 + 65)
=> Sum of 10 terms of the AP = 5 x 76 = 380

Solution : A = Arithmetic Mean = \frac{(4 + 6)}{2} = 5
G = Geometric Mean=\sqrt{4\times 6}= 4.8989
H = Harmonic Mean = \frac{(2 \times 4 \times  6) }{ (4 + 6)} = \frac{48 }{ 10} = 4.8
Therefore, A ≥ G ≥ H

First term, a = 32
Common ratio, r = \frac{16}{32} = \frac{8}{16} = \frac{4}{8} =\frac{1}{2} = 0.5
We know that for an infinite GP, Sum of terms =\frac{a} {(1 – r)}
=> Sum of terms of the GP =\frac{32}{ (1 – 0.5)} = \frac{32}{0.5} = 64

 Let the numbers be \frac{a}{r}, a, ar
=> (\frac{a}{r}) + a + a r = 26
=> \frac{a (1 + r + r^{2}) }{r} = 26
Also, it is given that product = 216
=> (\frac{a}{r}) \times  (a) \times  (a r) = 216
=> a³ = 216
=> a = 6
=> 6 \frac{(1 + r + r^{2})}{r} = 26
=> \frac{(1 + r + r^{2})}{ r} = \frac{26}{6} = \frac{13}{3}
=> 3 + 3 r + 3 r² = 13 r
=> 3 r² – 10 r + 3 = 0
=> (r – 3) (r – (\frac{1}{3}) ) = 0
=> r = 3 or r = \frac{1}{3}
Thus, the required numbers are 2, 6 and 18.

Solution :We know nth term of AP is [a + (n-1)d]
So, 7th term = a + 6d
and 11th term = a + 10d
Given a + 6d = 37 ………(1)
a + 10d = 57……….(2)
Subtract (1) form (2)
4d = 20
d = 5
put value of d in (1)
a + 6×5 = 37
a = 7
Hence, 1st term of the AP is 7.

Let the three consecutive parts of AP are (a-d), a, (a+d).
Given that
(a-d) + a + (a+d) = 21
3a = 21
a = 7
Again, (a-d)² + a² + (a+d)² = 155
a² + d² – 2ad + a² + a² + d² + 2ad = 155
3a² + 2d² = 155
put value of a
3(7)² + 2d² = 155
2d² = 155 – 147
d² = 4
d = ∓2
Hence, the largest part is (a+d) = 7+2 = 9

The series have multiple starting from 201, 204, ……..498
It becomes an AP having first term 210 and common difference 3.
Total number of natural numbers= [\frac{(\text{last term} – \text{first term})}{diff}] + 1
= [\frac{(498 – 201)}{3}] + 1
= \frac{297}{3} + 1
= 100

We know nth term of GP = ar^n-1
8th term = ar⁷ and 4th term =ar³
Given that
ar⁷ = 16ar³
=> r⁴ = 16
=> r = 2
Given ar⁵ = 64
put value of r above
a(2)⁵ = 64
a = \frac{64}{32} = 2
Hence, first term of the GP is 2

AM is 61 means there sum is 2×61 = 122
and, GM is 11 means there product is 11² = 121
Only possible values for A and B is 121 and 1.
So, value of A is 121.