the sum of the second and the fifth term of AP is 39 and the sum of the 6th and 8th term is 72 find the first three terms of AP
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Answer:
Step-by-step explanation:
a2+a5=39-----.1
a6+a8=72-----.2
in first equation ; a2+a5=39
a2=a+d ; a5=a+4d
add both RHS ; a+d+a+4d=39
2a+5d=39------.3
in second equation; a6+a8=72
a6=a+5d ; a8=a+7d
add both RHS ;a+5d+a+7d=72
2a+12d=72--------.4
now; subtract equations 4and 3
2a+12d=72
2a+5d=39
(-) (-) (-)
= (7d=33)
now; d=33/7
now substitute value of d in any one of the equation either 3 or 4
2a+5d=39 . (substitute d value in place of d)
2a+5(33/7)=39
2a+165/7=39 ;now take LCM (LCM of 7 and 1 =7)
now multilpy 7 with 2a
2a× 7+165/7=39
14a+165/7=39 ( now take 7 to right hand side so it will become multiply with 39)
14a +165=39×7
14a+165=273 (now shift 165 to right hand side so it will get subtracted by 273)
14a=273-165
14a=108.( now shift 14 to right hand side it will divide 108 )
a=108/14
a=54/7
now we got a and d value (a=54/7,d=33/7)
now we can find terms
we need to find first 3 terms of an AP
therefore we need to find a,a2,a3
a= we know a value is 54/7 .therefore a=54/7
and a2 , a2 means a+d so a+d= 54/7+33/7=87/7 (remember friends here no need for taking LCM because denominators are same so directly we can add numerator) so a2=87/7
now a3 , we know a3=a=2d
so 54/7+2(33/7) =120/7
so a3 =120/7
now we got a,a2,a3
a=54/7
a2=87/7
a3=120/7
therefore first 3 terms of an AP are (54/7,87/7,120/7)......
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