find five consecutive terms in an A.P. whose sum is 50 and product of first and fourth term is 28
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Let the first 5 terms be a, a+d, a +2d, a+3d, a+4d.
a+(a+3d)=28. (given)
2a+3d=28
a=14-(3d/2)---------(1)
a+a+d+a+2d+a+3d+a+4d=50. (given)
5a+10d=50
a+2d=10.
From (1)
14 - 3d/2 +2d= 10
d/2= -4
d=-8
Put value of d in (1)
a=14+ (24/2)
a=26.
five consecutive terms in an A.P are 26, 18, 10, 2, -6.
a+(a+3d)=28. (given)
2a+3d=28
a=14-(3d/2)---------(1)
a+a+d+a+2d+a+3d+a+4d=50. (given)
5a+10d=50
a+2d=10.
From (1)
14 - 3d/2 +2d= 10
d/2= -4
d=-8
Put value of d in (1)
a=14+ (24/2)
a=26.
five consecutive terms in an A.P are 26, 18, 10, 2, -6.
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