find the middle term of the sequence formed by all three digit numbers which leave remainder 5 when divided by 7. Also find the sum ofa
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Three digit multiples of 7 = 105,112,.........994
Now,
A.T.Q.,...
Required AP is =98+5,105+5,112+5,...........,994+5.
now.,
a=103, d=7, An=999
An=a+(n-1)d
999=103+(n-1)7
999-103=(n-1)7
896÷7=(n-1)
128+1=n
n=129
since,no.of terms are odd so,
middle term of AP =(n+1)/2
=(129+1)/2
=(130)/2
=65
T65=103+(64)×7
=103+448
=551
So, middle term = 551.
hope you got it..☺️
Now,
A.T.Q.,...
Required AP is =98+5,105+5,112+5,...........,994+5.
now.,
a=103, d=7, An=999
An=a+(n-1)d
999=103+(n-1)7
999-103=(n-1)7
896÷7=(n-1)
128+1=n
n=129
since,no.of terms are odd so,
middle term of AP =(n+1)/2
=(129+1)/2
=(130)/2
=65
T65=103+(64)×7
=103+448
=551
So, middle term = 551.
hope you got it..☺️
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