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2016-03-21T10:51:29+05:30
AP=8,10,12....
a=8,d=10-8=2
as we know,an=a+(n-1) x d
a₆₀ =8+(60-1)2
=8+118=126
so,now we find the sum of n terms
s₆₀=60/2[2 x 8 +(60-1)2]
=30[16+118]
=4020
as we taken last ten terms =60-10=50
so,
s₅₀=50/2[2×8+(50-1)2]
=25[16+98]
=2850
now sum of last ten terms= s₆₀-s₅₀=4020=-2850 =1170

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2016-03-21T10:51:35+05:30
A=8
d=10-8=2
tn=a+(n-1)d
 T60= 8+(60-1)2
       =8+59*2
       =8+118
       =126
now,
sn=n/2(2a+(n-1)d)
 S60=60/2(2*8+(60-1)2)
        =30(16+118)
         =4020
S50=50/2(2*8+(50-1)2)
       =25(16+49*2)
        = 25(16+98)
        =2850
now sum of last terms=S60-S50=4020-2850= 1170
∴sum of last 10 terms is 1170
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I m posted aldy, if the nth term of an AP is. .........
And also the ratio of the sums of m and n terms of an AP is. .........
Plz ans quickly
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