find the sum of the following APs:1/15,1/12,1/10,.......,to 11 term
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517/480
Solution:
Sn= (n/2 [2a+(n-1)d])
Here n= no. of terms , a=1st term, d= common difference
Here
n=11, a=1/15
d= 1/12 - 1/15
d=1/60
Substituting the values we get
S11 = (11/2[2*(1/15)+(11-1)1/60])
S11=11/2(2/15 + 1/16)
S11= 11/2 * 47/240
S11= 517/480
Solution:
Sn= (n/2 [2a+(n-1)d])
Here n= no. of terms , a=1st term, d= common difference
Here
n=11, a=1/15
d= 1/12 - 1/15
d=1/60
Substituting the values we get
S11 = (11/2[2*(1/15)+(11-1)1/60])
S11=11/2(2/15 + 1/16)
S11= 11/2 * 47/240
S11= 517/480
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