For an AP Sm=20 and Sn=10 and n-m=1,then prove that n=10÷a where a=d.

1.	For an AP Sm=20 and Sn=10 and n-m=1,then prove that n=10÷a where a=d.
Answer» \\(I \\space think \\space in \\space question S_m = 10 \\space and \\space S_n = 20 \\)Ans. Given :\xa0\\(S_m = 10\\)\\(S_n = 20 \\space \\space \\space \\space ..... (1)\\)=> n-m = 1\xa0=> m = n-1\xa0The sum of m terms i.e (n-1) terms\xa0\\(S_{n-1} =S_m = 10 \\space \\space \\space \\space \\space \\space ...... (2)\\)=>\xa0\\(n^{th} term = S_n - S_{n-1}\\)=>\\(n^{th} term = 20 - 10 = 10\\)=>\xa0\\(a + (n-1)d = 10\\)=>\xa0\\(a + (n-1)a = 10 \\space \\space \\space \\space \\space [as \\space a =d ]\\)=>\xa0\\(a+ na -a = 10 \\)=>\xa0\\(na = 10 \\)=>\xa0\\(n = {10\\over a}\\)Hence proved

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