DXC Quants Series and Progression Quiz-4

$$a_{n} = a\times r^{n-1}$$

$$a_{4} = a\times r^{4-1} = 81 $$-------(1)

$$a_{7} = a\times r^{7-1} = 2187 $$-------(2)

 

Divide 2 by 1

$$ r^{3} = 27 $$

r = 3

a= 2

$$ r = \frac{4}{2} = 2$$

$$a_{n} = a\times r^{n-1}$$

$$ 128 = 2\times 2^{n-1}$$

$$ 2^{7} = 2^{n}$$

n= 7

a = 16

d = 20 - 16 =4

nth term = a +(n-1) d = 16 + (n-1) 4 = 12 +4n

According to question

12 +4n = 44

n = 8

a = 4

$$ r =\frac{8}{4} =2 $$

$$s_{n} = a \times \frac{r^{n} -1}{r-1}$$

$$s_{n} = 4 \times \frac{2^{5} -1}{2-1} = 124 $$

So None

a= 200

$$s_{n} = a \times \frac{r^{n} -1}{r-1}$$

$$s_{n} = 100 \times \frac{1.20^{7} -1}{1.2 - 1} = 1291.5904 = Approx 1292$$

The term b should be the arithmetic mean of a and c.

So 2b = a + c

A.M = $$ \frac{a+b}{2}$$

G.M = $$\sqrt{ab}$$

So A.M is greater than or equal to G.M

The 5th term is a + 4d = 10  ....(1)

The 15th term is a + 14d = 30 ...(2)

 

10d=20 ....(2)-(1)

d=2

putting d in equation 1.

a=2

 

25th term= a+24d= $$ 2+ 24\times 2 $$=50

2nd term = a+d

5th term= a+4d

sum of 2nd term and 5th term is = 2a+5d

 

3rd term = a+2d

4th term= a+3d

sum of 3rd term and 4th term is = 2a+5d = sum of 2nd term and 5th term = 10

 

1st term = a

5th term= a+4d

sum of 1st term and 5th term is = 2a+4d

Average of 1st term and 5th term is= $${(2a+4d)}{2}$$ = a+2d = 3rd term.

 

Similarly , Average of 2nd term and 6th term is= 4rd term.

Common difference = difference of 2 consecutive term = 4th term - 3rd term = 100-80=20