find the middle term of sequence formed by all 3 digit no.which leave a remainder 3 . when divided by 4.also find the sum of all no. on both side of middle terms separately
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The A.P is
107 , 111 , 115 .... 999
a = 107
d = 4
an = 999
999 = 107 + (n-1)(4)
999 = 107 + 4n - 4
896 = 4n
n = 896/4 = 224
There are 224 terms
Middle term = 224/2 = 112th term
a112 = 107 + ( 112 - 1 )(4)
a112 = 107 + 444
a112 = 551
Sum of terms on left side
= 111/2 ( 2(107) + (111-1)(4) )
= 111/2 ( 214 + 440 )
= 2634
AP on left side
555,559...999
a = 555
no. of terms = 224 - 112 = 112
l = 999
sum = 112(555+999)/2 = 87024
Plz mark as brainliest
107 , 111 , 115 .... 999
a = 107
d = 4
an = 999
999 = 107 + (n-1)(4)
999 = 107 + 4n - 4
896 = 4n
n = 896/4 = 224
There are 224 terms
Middle term = 224/2 = 112th term
a112 = 107 + ( 112 - 1 )(4)
a112 = 107 + 444
a112 = 551
Sum of terms on left side
= 111/2 ( 2(107) + (111-1)(4) )
= 111/2 ( 214 + 440 )
= 2634
AP on left side
555,559...999
a = 555
no. of terms = 224 - 112 = 112
l = 999
sum = 112(555+999)/2 = 87024
Plz mark as brainliest
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