Find the A.P. whose first term is 100 and the sum of the first six terms is 5 times the sum of the next six terms.
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Answer:
Step-by-step explanation:
100+(100+d)+(100+2d)+(100+3d)+(100+4d)+(100+5d)=5((100+6d)+(100+7d)+(100+8d)+(100+9d)+(100+10d)+(100+11d))
600+15d=5(600+51d)
-240d=2400
d=-10
Now
T1=100
T2=90
T3=80........and so on
nidhi291:
it's not correct
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That what is answer?
it's right
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I can't understand properly
Send question again i will neetly explain with conditions
First term of AP,a=100
Let common difference =d
Sum of n terms of an A.P. =(n/2){2a+(n−1)d}
Sum of first 6 terms =(6/2)(2a+(6−1)d} =3(2a+5d)
Hence Sum of first 12 terms =(12/2)(2a+(12−1)d} =6(2a+11d)
Hence Sum of 7th to 12th terms=Sum of first 12 terms−=Sum of first 6 terms
=6(2a+11d)−3(2a+5d) =6a+51d
SInce sum of its 1st 6 terms is 5 times the sum of the next 6 terms
→3(2a+5d)=5(6a+51d)
→6a+15d =30a+255d
→6a−30a=255d−15d= 240d
−24a =240d
and d={24a/(−240)}
Since a= 100
→d=−24×100/240 =−10
Let common difference =d
Sum of n terms of an A.P. =(n/2){2a+(n−1)d}
Sum of first 6 terms =(6/2)(2a+(6−1)d} =3(2a+5d)
Hence Sum of first 12 terms =(12/2)(2a+(12−1)d} =6(2a+11d)
Hence Sum of 7th to 12th terms=Sum of first 12 terms−=Sum of first 6 terms
=6(2a+11d)−3(2a+5d) =6a+51d
SInce sum of its 1st 6 terms is 5 times the sum of the next 6 terms
→3(2a+5d)=5(6a+51d)
→6a+15d =30a+255d
→6a−30a=255d−15d= 240d
−24a =240d
and d={24a/(−240)}
Since a= 100
→d=−24×100/240 =−10
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