prateek wrote a list of three digit numbers which are divisible by 7 in increasing order the sum of those numbers is 1540 find the last number he wrote

frst 3 digit no. divisible by 7=105
so a=105 nd d=7
Sn =1540
n/2 (2a+(n-1) d)=1540
n (2×105+(n-1) 7)= 1540×2
n (210+7n-7)= 3080
n (7n+203)=3080
7n^2 +203n-3080=0
taking 7 as common
n^2 +29n-440=0
n^2+40n-11n-440=0
n (n+40)- 11 (n+40)=0
so
no. of terms =11
a11 = a+10d =105+7×10= 175

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The last number he wrote is 175...
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plz giv the whole solution 
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thanx it was quite easy I wasn't able to understand the Q.thumbs-up 4rm my side @khyati 
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thnks :)
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