plz plz plz plz help me out please let me know what is answer plz
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7
The list of 3 digit numbers divisible by 7 are :
105, 112, ........994.
The above list forms an AP with first term, a = 105 and common difference, d = 7
Let an = 994
so, a + (n - 1)d = 994
⇒ 105 + (n - 1)7 = 994
⇒ n = 128
Now, sum of all the 3 digit numbers divisible by 7 is given by,
S128 = (128/2) [2×105 +(128 - 1)7]
= 64 [210 + 889] = 64 × 1099 = 70336
105, 112, ........994.
The above list forms an AP with first term, a = 105 and common difference, d = 7
Let an = 994
so, a + (n - 1)d = 994
⇒ 105 + (n - 1)7 = 994
⇒ n = 128
Now, sum of all the 3 digit numbers divisible by 7 is given by,
S128 = (128/2) [2×105 +(128 - 1)7]
= 64 [210 + 889] = 64 × 1099 = 70336
aarushi47:
pls solve my first sum plz
Answered by
1
a = 105
d = 7
L = 994
=> a + (n-1) d = 994
=> 105 + (n-1)(7) = 994
=> (n-1) 7 = 889
=> n - 1 = 127
=> n = 128
Sn = 128 /2 [ 105 + 994]
= 64 ( 1099)
= 70336
d = 7
L = 994
=> a + (n-1) d = 994
=> 105 + (n-1)(7) = 994
=> (n-1) 7 = 889
=> n - 1 = 127
=> n = 128
Sn = 128 /2 [ 105 + 994]
= 64 ( 1099)
= 70336
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