Multiples of 9 lying between 300 and 700
Answers
Answered by
293
Here series will be = 306, 315, 324, 333,...........693
So we have the first term a = 306 and d = 9 and last term l = 693
693 = 306 + (n-1)9
77 = 34 + (n-1)
n = 77 - 34 +1
= 44
Sum = n/2(a+l)
= 44/2(306+693)
= 22(999)
= 21978.
Hope this helps!
So we have the first term a = 306 and d = 9 and last term l = 693
693 = 306 + (n-1)9
77 = 34 + (n-1)
n = 77 - 34 +1
= 44
Sum = n/2(a+l)
= 44/2(306+693)
= 22(999)
= 21978.
Hope this helps!
Answered by
75
HEY THERE!!!
Question;-
Find the sum of all multiples of 9 lying between 300 and 700.
Method of Solution;-
Firstly, find which number lies on sum of all multiples of 9 lying between 300 and 700.
Number which are multiples of 9 lying between 300 and 700 Given Below in the form of Arithmetic Sequence or Progression.
306, 315, 324, 333, ..., 693.
here,
a = 306, d = (315 - 306) = 9 and l = 693.
Let the number of terms be n.
Then Tn = 693
⇒ a + (n - 1)d = 693
= 306 + (n - 1) 9 = 693
= 9n = 396
= n = 44
∴ Required sum = n /2(a+l)
= 44/2(306+693)
=22(306+693)
=22(999)
=21978
Hence, sum of all multiples of 9 lying between 300 and 700 = 21,978.
Question;-
Find the sum of all multiples of 9 lying between 300 and 700.
Method of Solution;-
Firstly, find which number lies on sum of all multiples of 9 lying between 300 and 700.
Number which are multiples of 9 lying between 300 and 700 Given Below in the form of Arithmetic Sequence or Progression.
306, 315, 324, 333, ..., 693.
here,
a = 306, d = (315 - 306) = 9 and l = 693.
Let the number of terms be n.
Then Tn = 693
⇒ a + (n - 1)d = 693
= 306 + (n - 1) 9 = 693
= 9n = 396
= n = 44
∴ Required sum = n /2(a+l)
= 44/2(306+693)
=22(306+693)
=22(999)
=21978
Hence, sum of all multiples of 9 lying between 300 and 700 = 21,978.
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