hom many terms of the AP -6,-11/2,-5,..... are needed to give the sum -25?
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As we know that Sum of n terms of an A.P. is given by
S = n/2 {2a +(n-1)d}
where
n = no. of terms
a = first term
d = common difference
S = sum of n terms
Now,By putting the values in the formula
We get
-25 = n/2 {2*(-6)+(n-1)(1/2)} ........... equation(i)
-50= n{-12+0.5n-0.5}
-50=n{-12.5+0.5n}
-50 = -12.5n +0.5n^2
0.5n^2-12.5n+50 = 0
By solving the above equation ,
n = 20, 5.
Again putting the value of n in the equation (i),
We find n=5 as the no. of terms giving a sum of -25.
S = n/2 {2a +(n-1)d}
where
n = no. of terms
a = first term
d = common difference
S = sum of n terms
Now,By putting the values in the formula
We get
-25 = n/2 {2*(-6)+(n-1)(1/2)} ........... equation(i)
-50= n{-12+0.5n-0.5}
-50=n{-12.5+0.5n}
-50 = -12.5n +0.5n^2
0.5n^2-12.5n+50 = 0
By solving the above equation ,
n = 20, 5.
Again putting the value of n in the equation (i),
We find n=5 as the no. of terms giving a sum of -25.
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