Zoho Progression Quiz-1

nth  term of A.P = a+(n-1)d

                          = 3 + (25 - 1) 4

                          = 3 + 24*4

                          = 3 + 96

                          = 99

Explanation - 

a+11d = 6 —---- (1)

a + 7d = 4 ______(2)

 

From eq 1 and 2 we get - 

d = ½

 

Put d =½ in eq 1

 

a + 11 (1/2) = 6

a + 5.5 = 6

a = 0.5 = ½

S_{n} = \frac{n}{2} \times (2a+(n-1)d)

S_{20} = \frac{20}{2} (2*½+(20 -1) 1/2)

S_{20} = 10 (1 + (19)*(\frac{1}{2})

S_{20} = 10 (1 + 9.5)

S_{20} = 10(10.5)

S_{20} = 105

= \left ( 2^{2} +4^{2}+6^{2}+ ....+20^{2}\right ) = \left ( 1*2^{2} +2*2^{2}+2*3^{2}+ ....+2*10^{2}\right )

= 22*12 + 22*22+ 22*32+.....22*102

= 22[12+22+......102]

= 4*385

= 1540

Given : a+(a+d)+(a+2d) = 60

 3a + 3d = 60 —--1

The middle terms are 20th, 21st, 22nd terms

=>  a+19d+a+20d+a+21d = 120

=> 3a + 60d = 120 —-- 2

=> from eq 1 and 2

We get - d = 20/19

Put d= 20/19 in eq 1

3a + 3 (20/19) = 60

3a + 3*1.52 =60

3a + 4.56 = 60

3a =  60 - 4.56 

3a = 55.44

a = 18.48

Given = 1+ 2 + 3 +......50

=> sum of first n natural numbers = n(n+1)/2

Where n is number of term

Sum = n(n+1)/2 

Sum = 50 (50+1)/2

Sum = (50 * 51) /2

Sum = 2550 / 2

Sum = 1275

 

Hence 1275 is only divisible by 51.

Here a =2, r = 4/2 =2, 

Let sum of terms be n

Then, tn = 2048

ar^{n-1} = 2048

2*2^{n-1}= 2048

2^{n-1} = 1024

2^{n-1} = 2^10

n = 10 +1

n = 11

 

So there are a total of 11 terms in the given G.P.

Let us consider the angles are - a, ar, 2ar, 3ar

Sum  of  the angles = a ( r 4- 1 ) /r -1 = a ( r 2 + 1 ) ( r + 1 ) = 360

a< 90 , and  ar< 90,  Therefore,  a ( 1 + r ) <  180,  or   ( r 2 + 1 ) > 2

Therefore, r  is  not  equal  to  1.  Trying  for  r  =  2  we  get  a  = 24  Therefore, The  angles  are  24, 48, 96  and  192.

The required numbers are : 9, 18, 27, 36, ………. 99

This is an A.P with a = 9 d = 9 

Let us consider the series having n term

Then, tn = 99

tn = a + (n -1)d

99 = 9 + (n -1) 9

9 + (n-1)9 = 99

9n - 9 = 90

9n = 99

n =11

 

Therefore there are 11 numbers between 7 to 100 which are divisible by 9.

We have -

Sn = n/2[2a+(n-1)d]

Tn = a+(n-1)d

\frac{S_{1}}{S_{6}} = \frac{1}{18}

\frac{1}{18} = \frac{a}{\frac{n}{2} [2a+(n-1)d]}

\frac{1}{18} = \frac{a}{\frac{6}{2} [2a + 5d]}

\frac{1}{18} = \frac{a}{3 [2a + 5d]}

After comparing a = 1

Therefore, 18 = 3(2a+5d)

18 = 3 (2 + 5d)

2 + 5d = 6

5d = 4

d = \frac{4}{5}

\frac{T_{1}}{T_{3}} = \frac{a}{a+2d} = \frac{a}{a+\frac{8}{5}}

Put a = 1

Hence, \frac{T_{1}}{T_{3}} = \frac{5}{13}

 

Let the numbers be - a-d, a, a+d

Then, a - d + a + a + d = 27

3a = 27

a  = 9

And (a-d)(a+d) = 97

a^{2}-d^{2} = 97

d^{2} = 97 - 9^{2}

-d^{2} = 97 - 81

 

-d^{2} = - 16

 

d = 4

 

Hence, the numbers are 5, 13, and 9 when d = -4, In both cases numbers, are the same.