Question

Given the sum s20= 910 of an arithmetic sequence and A20 = 95, Find A1​

Answer 1

Step-by-step explanation:

Sn=n/2(2a+(n-1)d)

910=n/2{2a+(n-1)d}

1820=n{2a+(n-1)d}

1820=20{2a+(20-1)d}

1820=20{2a+19d}

91=2a+19d

An=a+(n-1)d

95=a+(20-1)d

95=a+19d

91=2a+19d

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4=-a

a=-4

91=2a+19d

91=2(-4)+19d

91=-8+19d

91+8=19d

99=19d

d=99/19

A=-4