We have to consider the given sum as S_{n}. Instead of S_{n}, we may use one of the formulas given below.
S_{n} = (n/2) [a + l] (or)
S_{n} = (n/2) [2a + (n - 1)d]
a = first term, d = common difference and n = number of terms.
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 :
t_{3} = 7
t_{7} = 3t_{3}+2
t_{7} = 3(7)+2
t_{7} = 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]
S_{20} = (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
S_{11} = 44
(11/2)[2a+(11-1)d] = 44
2a+10d = 44 ⋅ (2/11)
2a+10d = 8 ----- (1)
Sum of next 11 term = 55
S_{22} = S_{11} + 55
S_{22} = 44 + 55
S_{22 }= 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) + ............
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