Arithmetic Progressions 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