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| 1. |
In A.P if 1st S10=-80 & sum of nxt ten ter=-280...find A.P |
| Answer» Let A.P be a , a + d, a + 2d, ...a is the first term, d is the common differenceThe formula of sum of n terms is\xa0{tex}S_{n}=\\frac{n}{2}[2 a+(n-1) d]{/tex}Where n is the number of termsThe sum of its first 10 term is - 80. Therefore,{tex}-80=\\frac{10}{2}[2 a+(10-1) d]{/tex}-80 = 5[2a + 9d]-16 = 2a + 9d ...(1)Sum of its next 10 terms is -280.{tex}S_{20}=\\frac{20}{2}[2 a+(20-1) d]{/tex}{tex}S_{20}{/tex}\xa0= 10[2a + 19d]Since sum of the next 10 terms of AP is -280. Therefore,{tex}S_{20}-S_{10}{/tex}\xa0=-28010[2a + 19d] - (-80) = -28010[2a + 19d] = 3602a + 19d = -36 ...(2)Subtract (1) from (2)2a + 19d - 2a - 9d = -36 + 1610d = -20d = -2Put value of d in (1)-16 = 2a + 9d-16 = 2a + 9(-2)-16 = 2a - 182 = 2a\xa0a = 1Therefore, The AP series is 1, -1, -3, -5 ... | |