The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP.
Answers
Answer:
Answer:given
Answer:givena4+a8=24
Answer:givena4+a8=24or , a+3d +a+7d=24
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - -
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq1
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=24
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=24
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-50
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13)
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8 (a2= -8)
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8 (a2= -8)
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8 (a2= -8) a3= a+2d = -13+2×5= -13+10 = -3
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8 (a2= -8) a3= a+2d = -13+2×5= -13+10 = -3 (a3= -3)
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8 (a2= -8) a3= a+2d = -13+2×5= -13+10 = -3 (a3= -3)thus first three terms = -13, -8, -3
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8 (a2= -8) a3= a+2d = -13+2×5= -13+10 = -3 (a3= -3)thus first three terms = -13, -8, -3I hope this is helpful.
Answer:givena4+a8=24or , a+3d +a+7d=24 2a+10d=24...eq1a6+a10=44or, 2a+14d=44....eq2from eq 1 and 2 2a+14d=44 2a+10d=24 - - - 4d=20 (d=5).......put the value of d in eq12a+10×(5)=242a+50=242a=24-502a= -26(a= -13)now , (a1= -13) a2= a+d = -13+5= -8 (a2= -8) a3= a+2d = -13+2×5= -13+10 = -3 (a3= -3)thus first three terms = -13, -8, -3I hope this is helpful.thank you
Answer:
-13,-8
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