已知等差数列｛an｝的前n项和sn,a2+a7=-23,s10=-145

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已知等差数列｛an｝的前n项和sn,a2+a7=-23,s10=-145
(1)求数列｛an｝的通项公式
(2)设数列{bn/an}是首项为1,公比为c的等比数列,求数列{bn}的前n项和Tn

1
因为等差数列
a2+a7 +2d = a2 +a9 = a1+ a10
s10= 5*(a1 +a10 ) = 5(a2+a7 +2d) = -23*5 +5d
-23*5 +10d = -145
10d = -145 +115
d=-3
a2+a7 = a1+d + a1+6d = 2a1 +7d
2a1 + 7 * (-3) = -23
a1 = -1
数列｛an｝的通项公式 an = -1 + (n-1) (-3) = -3n +2
2
bn/an = 1 *C ^(n-1)
bn = C^(n-1) * (-3n +2)
b1 = C^0 * (-3*1 +2) = -1
求数列{bn}的前n项和Tn
CTn - Tn = C(b1+b2+...+bn) -(b1+b2+...+bn)
= -b1 + (-d) (C + C^2 + ...+ C^(n-1)) + C bn
= -(-1) + 3*(C-C^n)/(1-C) +C^n* (-3n +2)
= (-3n +2)*C^n + 3*(C-C^n)/(1-C) +1

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