THE 10TH AND 18TH TERMS OF AN AP ARE 41 AND 73. FIND THE 45TH TERM?
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1
Let first term be 'a' nd common difference be 'd'
now,
a+9d = 41 ------- (i)
a+17d = 73 ------(ii)
on subtracting both equation we get,
a+9d - (a+17d) = 41 - 73
a+9d-a-17d = -32
-8d = -32
d =4
putting value of 'd' in (i) we get ,
a + 9*4 = 41
a + 36 = 41
a = 41-36
a = 5
therefore, 45th term is
a+44d = 5 + 44*4 = 181
now,
a+9d = 41 ------- (i)
a+17d = 73 ------(ii)
on subtracting both equation we get,
a+9d - (a+17d) = 41 - 73
a+9d-a-17d = -32
-8d = -32
d =4
putting value of 'd' in (i) we get ,
a + 9*4 = 41
a + 36 = 41
a = 41-36
a = 5
therefore, 45th term is
a+44d = 5 + 44*4 = 181
Sriharan2002:
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SIR HOW DID YOU GET THE a=? answer.... please explain
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Answered by
0
d=Tp-Tq/p-q
d=T10-T18/10-18
d= 41-73/10-18
d= -32/-8= 4
therefore T10=a+9d
41=a+9.4
a=5
T45= a+44d
5+44.4=5+176=181( substitute the a and d values)
Therefore 181 is the 45th term.
d=T10-T18/10-18
d= 41-73/10-18
d= -32/-8= 4
therefore T10=a+9d
41=a+9.4
a=5
T45= a+44d
5+44.4=5+176=181( substitute the a and d values)
Therefore 181 is the 45th term.
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