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Question Number 188360 by 073 last updated on 28/Feb/23

Answered by Rasheed.Sindhi last updated on 28/Feb/23

$${\mathrm{S}}_{\mathrm{N}}=\frac{\mathrm{N}}{2}(a+l)$$$$\mathrm{A}=\frac{n-2}{2}(1+n-2)=\frac{(n-2)(n-1)}{2}$$$$\mathrm{B}=\frac{n-14}{2}(15+n)=\frac{(n-14)(n+15)}{2}$$$$\mathrm{A}-\mathrm{B}=\frac{(n-2)(n-1)}{2}-\frac{(n-14)(n+15)}{2}=42$$$$=\frac{(n-2)(n-1)-(n-14)(n+15)}{2}=42$$$$=\frac{(n^2-3n+2)-(n^2+n-210)}{2}=42$$$$=\frac{-4n+212}{2}=106-2n=42$$$$n=32$$

Commented by 073 last updated on 01/Mar/23

$$\mathrm{thanks}\mathrm{sir}$$

Answered by manxsol last updated on 28/Feb/23

$$A-B=\sum _1^{14}n-(n-1)-n$$$$42=\frac{14\times 15}{2}-2n+1$$$$42=105-2n+1$$$$2n=64$$$$n=32$$$$$$$$$$$$$$

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