Arithmetic Progression Questions and Answers

As we know about the nth term of AP is [a + (n-1)d]
So, 6th term = a + 5d
and 10th term = a + 9d
Given a + 5d = 36 ………(1)
a + 9d = 56……….(2)
Subtract (1) form (2)
4d = 20
d = 5
put value of d in (1)
a + 5×5 = 36
a =11
Hence, 1st term of the AP is 11.

Let the three consecutive parts of AP are (a-d), a, (a+d).
Given that
(a-d) + a + (a+d) = 24
3a = 24
a = 8
Again, (a-d)² + a² + (a+d)² = 208
a² + d² – 2ad + a² + a² + d² + 2ad = 208
3a² + 2d² = 208
put value of a
3(8)² + 2d² = 208
2d² = 208 – 192
d² = 8
d = ∓2.9
Hence, the largest part is (a+d) = 8+2.9 = 10.9

The series have multiple starting from 320, 360, ……..480
It becomes an AP having common difference 40.
Total number of natural numbers =

n=  [ \frac{(l-a)}{d} ]+ 1

= [\frac{480 – 320}{40}] + 1
= \frac{160}{40} + 1 = 5

n^th term = a + (n-1)d, 33 = 7 + (n-1)3,

n = \frac{29}{3}

n = 9.666

Clearly 9.66 is not an integer. So 33 is not a term in this series.

n^th term = a + (n-1)d

44 = 17 + (n-1)3,

n = \frac{30}{3} = 10

n = 10

Clearly 10 is an integer. So 44 is a term in this series.

It becomes an infinite sum of series.
So, use \frac{a}{(1-r)} to calculate the distance
Ball rebounds to \frac{3}{4} of its height -> 360\times (\frac{3}{4}) = 270
360 + \frac{270 \times 2}{1-(\frac{3}{4})}
= 360 + 2160
= 2520
Hence, the full distance of elastic toy=2520

It is an AP 2000, 3500, 5000, ……..so on.
As we know in total there 60 months in 5 years.
We need to calculate the 60th term of the series.
Common difference d = 1500
a₆₀ = a + (n-1)d
a₆₀ = 2000 + 59 x 1500
a₆₀ = 2000 + 88500 = 90500
After completion of 5 years of service his salary will be Rs 90500.

Difference is:

d= 8√5-9√5 = 7√5-8√5 = -√5

Therefore given arrangement is in Arithmetic progression.

Clearly the given series is in Arithmetic progression

2√4+√4+0+.....  is an A.P.

Now

a=2√4, d=−√4

. Hence 8^th term of the series

=2√4+(8−1)(−√4)=−5√4.

Given that 8th term = 0

8th term = a+(8−1)d = a+7d

so a+7d = 0

Now ratio of 28th and 18th terms

= \frac{a+27d}{a+17d} = \frac{(a+7d)+20d}{(a+7d)+10d}=\frac{20d}{10d} = \frac{2}{1} = 2:1

Sequence

f(n)=an+b; n∈N is an A.P. Putting n=1, 2, 3, 4, ..........,

we get the sequence (a+b), (2a+b), (3a+b),.........

which  is  an  A.P. Where first term (A)=(a+b) and common difference;d=a

Given sequence is

(−10+19i)(−8+16i)(−6+13i)(−4+10i)(−2+7i)(−0+4i)

Therefore 6th term is purely imaginary.

Since, given series is (1×3)+(3×5)+(5×7)+(7×9)+.......

In 1 , 3 , 5.....

nth term = 1+(n-1)2 = (2n-1)

In 3 , 5 , 7.....

nth term = 3+(n-1)2 = (2n+1)

Hence nth term of the series is (2n-1)(2n+1).

Given series 201+199+197+.........+147

So, first term a=201, difference d = −2 and last term, l = 147

As per the formula

T_{l}=a+(n−1)d

=>147=201+(n−1)(−2)

=>−54=(n−1)(−2)

n=28

nth term of A.P. =a+(n-1)d = 3+(17-1)4 = 67.

as per the formula:

(n\times \frac{(n+1)}{2})

\frac{(200\times 201)}{2}

= 100 x 201

201 is an odd number and 100 is divisible by 2.

So 100 x 201 will be divisible by 2,4 and 5.

Clearly the numbers which will be divisible by 8:

16, 24,32, ………………. 88

This is an Arithmetic Progression where a = 16 and common difference = 8

Thus: T_{n} = 88

or a+(n-1)d=88

or 16+(n-1)8=88

or 16+8n-8=88;

8n=80

n = 10

Assume the numbers as:

a-d, a, a+d

or a-d+a+a+d=36

or 3a=36

a=12

now (a-d) (a+d)=80

a²-d²=80

144-d²=80

d²=64

d=8

thus the numbers will be: 4, 12 and 20

Here, a = 3, d = 11 – 3 = 8 and n = 11.

We have a_{n} = a + (n – 1) d

= 3 + (11 – 1) × 8 = 3 + 80 = 83 Therefore, the 11th term of the given AP is 83.

Here, a = 11, d = 7 – 11 = – 4, l = – 65, where l = a + (n – 1) d

but as per the information given in the question If we write the given AP in the reverse order, then a = – 65 and d = 4. So, we have to find the 10^th term with new a and d.

so = – 65 + (10 – 1) × 4 = – 65+ 36 = – 29

So, the 10th term will be – 29.

Clearly the above mentioned series forms an AP series.

Assume the total number of rows as n.

Then a = 25, d = 23 – 25 = – 2,

a_{n} = a + (n – 1) d We have, 5 = 25 + (n – 1)(– 2)

i.e., – 20 = (n – 1)(– 2) i.e., n = 11 So, there are total 11 rows.

The nth term is an=3+2n

so by using this we can find the value of terms like a_{1} = 5,a_{2}  = 7,a_{3} = 9

and a_{27}=57

now the first term is 5 and the last term is 57 and the common difference is 2

By applying the formula we will get:

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

so S_{27} = \frac{27}{2}(10+26\times 2)

= 837

n^th term = first term + (number of terms – 1) common difference.

60 = 15 + (n-1)3,

60=15+3n-3=

48=3n

n = 16

n^th term = a + (n – 1)d,

n^th term = 4+(12-1)4

= 4+44

= 48

d = 147 - 150 = -3

Assume n^th term=0 , the following term will be first negative term.
0 = 150 + (n-1) – 3,

0 = 150-3n+3

3n=153

n = 51

So at n = 51 + 1 =52 the first negative term would occur.

A_{n} = 5 + 3n is a linear expression. This contains AP with difference of 3.

The term y should be the arithmetic mean of term x and z.

Let a-d , a ,a+d three terms in AP

Sum = 180

a-d+a+a+d = 180

a = 60

Mid term = a = 60

n^th term = a + (n-1)d,

26 = 6 + (n-1)3,

n = \frac{23}{3}

n = 7.666

Clearly 7.66 is not an integer. So 26 is not a term in this series.

In A.P. the difference in any term with previous term is same.