# 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

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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