FIND THE SUM OF ARITHMETIC SERIES WITH GIVEN DESCRIPTION

Problem 1 :

Find the sum of all 3 digit natural numbers, which are divisible by 9.

Solution :

3 digit number starts from 100 and ends with 999. From this sequence we have to find number of terms which are divisible by 9 and also we have to find their sum.

  108, 117, 126,........................999

The first number which is divisible by 9 from this sequence is 108, the second number will be 117 and the last number

will be 999.

108 + 117+ 126 + ........... + 999

a = 108, d = 117 - 108 = 9 and l = 999      

to find number of terms, we have to use the formula for (n)

  n  =  [(l-a)/d]+1

  n  =  [(999-108)/9] + 1

  n  =  [891/9] + 1

  n  =  99 + 1

  n  =  100

 S_n  =  (n/2)[a+l]

= (100/2)[108+999]

=  50 (1107)

=  55350

Problem 2 :

Find the sum of first 20 terms of the arithmetic series in which 3^rd term is 7 and 7^th term is 2 more than three times its 3^rd term.

Solution :

t₃  =  7

t₇  =  3t₃+2

t₇  =  3(7)+2

t₇  =  23

a+2d  =  7   -----(1)

a+6d  =  23 -----(2)

(1) - (2)

-4d  =  -16

d  =  4

By applying the value of d in (1), we get

a = -1

Now we need to find sum of 20 terms

S_n  =  (n/2)[2a+(n-1)d]

S₂₀  =  (20/2) [2(-1)+(20-1)(4) ]

=  10 [-2+19(4)]

=  10 (-2+76)

=  740

Problem 3 :

In an arithmetic series, the sum of first 11 terms is 44 and the that of the next 11 terms is 55.

Find the arithmetic series.

Solution :

Sum of first 11 terms  =  44

S₁₁  =  44

(11/2)[2a+(11-1)d]  =  44

  2a+10d  =  44 ⋅ (2/11)

  2a+10d  =  8   ----- (1)

Sum of next 11 term  =  55

  S₂₂  =  S₁₁ + 55

S₂₂  =  44 + 55

S₂₂  =  99

(22/2)[2a+(22-1)d]  =  99

2a+21d  =  99/11

2a+21d  =  9   ----- (2)

(1) - (2)

-11d  =  -1

  d = 1/11

By applying the value of d = 1/11 in (1), we get

  2a+10(1/11)  =  8

  2a+10/11  =  8

  2a  =  8-(10/11) ==> 78/11 ==> a = 39/11

Therefore the series is

(39/11) + (40/11) + (41/11) + ............

Problem 4 :

If the sum (Sn) of n terms of an Arithmetic progression is (2n² + n). What is the difference of its 10^th and 1^st term ?

a)  207     b)  36    c)  90     d)  63

Solution :

To find the nth term of the arithmetic progression, we have to find the difference between sum of n - 1 terms and sum of n terms.

S_n = 2n² + n

Sum of n - 1 terms :

S_n-1 = 2(n-1)² + n - 1

= 2(n² - 2n + 1) + n - 1

= 2n² - 4n + 2 + n - 1

= 2n² - 3n + 1

Sn - S_n-1 = (2n² + n) - (2n² - 3n + 1)

= 2n² + n - 2n² + 3n - 1

tn = 4n - 1

Difference = 39 - 3

= 36

Problem 5 :

The sum of all two digit odd numbers is 

a)  2475      b)  2575       c)  4950      d)  5049

Solution :

The first two digit number is 10, but it is not an even number.

11, 13, 15, ................99

Number of terms in the above sequence :

t_n = a + (n - 1)d

a = 11, d = 13 - 11 ==> 2, tn = 99

99 = 11 + (n - 1) 2

99 - 11 = 2(n - 1)

88/2 = n - 1

44 = n - 1

n = 44 + 1

n = 45

There are 45 terms.

Sn = (n/2)[2a+(n - 1)d] 

= (45/2)[2(11) + (45 - 1)2]

= (45/2) [ 22 + 44(2)]

= (45/2) [ 22 + 88]

= (45/2) (110)

= 45(55)

= 2475

So, the required sum is 2475.

Problem 6 :

If each month $100 increases in any sum then find out the total sum after 10 months, if the sum of the first month is $2000.

a)  $24500    b)  $24000     c)  $50000    d)  $60000

Solution :

First term (a) = 2000

Common difference (d) = 100

The sum after 10 months :

Sn = (n/2)[2a+(n - 1)d] 

= (10/2)[2(2000) + (10 - 1) 100]

= 5 [4000 + 9(100)]

= 5[4000 + 900]

= 5(4900)

= 24500

So, the required sum is 24500.

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