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प्रश्न
Q.12
बेरीज
उत्तर
`"Sum of first 10 terms" = -80`
`⇒ 10/2[2a+(10-1)d]=-80`
`⇒5[2a+9d]=-80`
`⇒2a+9d=-16............(1)`
And, sum of next 10 term =-280
`"⇒ Sum of first 20 terms= sun of first ten terms +sun of nect 10 terms "`
`=-80+(-280)`
` =-360`
`⇒20/2[2a+(20-1)d]=-360`
`⇒10[2a+19d0=-360]`
`⇒ 2a+19d=-36 .............(2)`
Substracting (1) from (2), we get
`10d=-20⇒ d=-2`
`⇒2a+9xx(-2)=-16`
`⇒2a-18=-16`
`⇒2a=2⇒a=1`
∴ Required A.P = `a,a++d, a+2d, a+3d,..........`
` = 1,-1,-3,-5,..................`
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