An A.P consists of 57 terms of which 7th term is 13 and the last term is 108.Find the 45th term of this AP
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given,
n= 57
a₇ =13 = a+6d → (1)
a₅₇ = 108 = a+56d → (2)
a₄₅ = a+44d
we know the formula for nth term is,
an = a+(n-1)d
a = first term
d = common diff
n= no. of terms
subtracting (1) and (2) we get,
a+56d = 108
a+6d = 13
⇒ d = 1.9
Substituting the value of d in 1,
a+6*1.9 =13
⇒ a = 1.6
∴45th term = a+44d
⇒1.6+(44*1.9) = 85.2
on rounding off 85
hope it helps.... :)
n= 57
a₇ =13 = a+6d → (1)
a₅₇ = 108 = a+56d → (2)
a₄₅ = a+44d
we know the formula for nth term is,
an = a+(n-1)d
a = first term
d = common diff
n= no. of terms
subtracting (1) and (2) we get,
a+56d = 108
a+6d = 13
⇒ d = 1.9
Substituting the value of d in 1,
a+6*1.9 =13
⇒ a = 1.6
∴45th term = a+44d
⇒1.6+(44*1.9) = 85.2
on rounding off 85
hope it helps.... :)
niyamee:
kindly mrk it as brainliest.....
thank u .. :)
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can u help me in other three questions?please
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the sum of first 7terms of an AP is 49 and that of first 17 terms of it is 289. find the sum of first n terms
can u pls post it as qstn so tht i can use the symbols properly....
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ok
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