The fourth term of an AP is 10.if the eleventh term is one more three times of theach fourth term,find the sumo it's 25 terms
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sum of 25 terms is 925
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a + 9d= 10......(1)
a+ 10d = 3(a+3d)+1.......(2)
a+10d = 3a+9d+1
10d–9d= 3a–a +1
[d= 2a +1].........(3)
Substituting value of d in eqn.(1)
a+ 9(2a+1)= 10
a+ 18a+9= 10
19a= 1
[a=1/19]
putting value of a in eqn (3),
d= 2a+1
d= 2×1/19+1
d= 2/19+1
d= 2+19
[d= 21].
n= 25
Sn= n/2(2a+ (n–1)d)
Sn= 25/2(2/19+ (24×21))
Sn= 12.5(2/19+504)
Sn= 12.5(0.15+504)
Sn= 12.5×504.15
Sn= 6301.88
or
Sn= 6302.[Answer]
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a+ 10d = 3(a+3d)+1.......(2)
a+10d = 3a+9d+1
10d–9d= 3a–a +1
[d= 2a +1].........(3)
Substituting value of d in eqn.(1)
a+ 9(2a+1)= 10
a+ 18a+9= 10
19a= 1
[a=1/19]
putting value of a in eqn (3),
d= 2a+1
d= 2×1/19+1
d= 2/19+1
d= 2+19
[d= 21].
n= 25
Sn= n/2(2a+ (n–1)d)
Sn= 25/2(2/19+ (24×21))
Sn= 12.5(2/19+504)
Sn= 12.5(0.15+504)
Sn= 12.5×504.15
Sn= 6301.88
or
Sn= 6302.[Answer]
Hope it helps you
Mark it as brainleist.
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