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In an AP, the sum of first ten terms is -150 and the sum of its next ten terms is -550.Find the AP |
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Answer» Given:Sum of first ten terms of an AP = -150Sum of next ten terms of an AP =-550To find:APSolution:Sum of n terms of an AP Sum of first 10 terms -150 = 5(2a+9d)-150=10a+45d…..(i)Sum of next 10 terms -150-550=10(2a+19d)-700=10(2a+19d)-70=2a+19d …….(ii)Combining 1 and 2, we getMultiplying (ii) by 5-350=10a+95d….(iii)Subtracting (i) from (iii) we get,-350+150=10a+95d-10a-45d-200=50dd=-4substitute d=-4 in equation (i) we get-150=10a+45d-150=10a+45(-4)-150=10a-18010a=-150+18010a=30a=3hence, a=3 and d=-4The AP will be a, a+d, a+2d, a+3d,…Then the AP will be 3, -1, -5, -9,…." Given:Sum of first ten terms of an AP = -150Sum of next ten terms of an AP =-550To find:APSolution:Sum of n terms of an AP Sum of first 10 terms -150 = 5(2a+9d)-150=10a+45d…..(i)Sum of next 10 terms -150-550=10(2a+19d)-700=10(2a+19d)-70=2a+19d …….(ii)Combining 1 and 2, we getMultiplying (ii) by 5-350=10a+95d….(iii)Subtracting (i) from (iii) we get,-350+150=10a+95d-10a-45d-200=50dd=-4substitute d=-4 in equation (i) we get-150=10a+45d-150=10a+45(-4)-150=10a-18010a=-150+18010a=30a=3hence, a=3 and d=-4The AP will be a, a+d, a+2d, a+3d,…Then the AP will be 3, -1, -5, -9,…." So .. Given that Sum of first 10 terms is -150 and of next 10 is -550 soS10 =-150-150 = 10/2(2a+(10-1)d=2a+9d = -30 .....(1) And again given that sum of next ten terms is -550So -150 +(-550)=-770-770= 20/2(2a+(20-1)d)=2a+19d=-70......(2)By applying elimination b/w bothWe get 10d = -40d= -4...Putting d=-4 in eq 1....2a+9(-4) = -302a-36 =-302a = 6 a = 3So Ap would be .... 3 , -1,-5,-9,..... Hope it will help you:)..... Have a nice day !! |
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