In an A.P. of 19th term is 52 and 38th term is 128 find sum of first 56 terms
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Answered by
5
a19=52
a+18d=52
a=52-18d
a38=128
a+37d=128
52-18d+37d=128
52+19d=128
19d=76
d=4
a=52-18*4
a=52-72
a=-20
a56=a+55d
-20+55*4
-20+220
=200 is the answer
a+18d=52
a=52-18d
a38=128
a+37d=128
52-18d+37d=128
52+19d=128
19d=76
d=4
a=52-18*4
a=52-72
a=-20
a56=a+55d
-20+55*4
-20+220
=200 is the answer
Answered by
5
T19=52
a+(n-1)d=52
a+(19-1)d=52
a+18d=52 -------(1)
T38=128
a+(n-1)d=128
a+(38-1)d=128
a+37d=128 -------(2)
subract equation (2) - (1)
a+37d-(a+18d)=128-52
19d=76
d=4
substitute in equation (1)
a+18d=52
a+18×4=52
a+72=52
a=(-20)
Sn=n/2(2a+(n-1)d)
S56=56/2(-40+(56-1)4)
S56=28×180
S56=5040
sum of 56 terms is 5040
a+(n-1)d=52
a+(19-1)d=52
a+18d=52 -------(1)
T38=128
a+(n-1)d=128
a+(38-1)d=128
a+37d=128 -------(2)
subract equation (2) - (1)
a+37d-(a+18d)=128-52
19d=76
d=4
substitute in equation (1)
a+18d=52
a+18×4=52
a+72=52
a=(-20)
Sn=n/2(2a+(n-1)d)
S56=56/2(-40+(56-1)4)
S56=28×180
S56=5040
sum of 56 terms is 5040
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