the sum of first 6 terms of an ap is 42 the ratio of 10th term and 13th term is 1 ratio 3 find the first term and 21st term
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you answer will be it may be wrong 2 and 8
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S6=n/2[2a+(n-1)d]
84=6[2a+5d]
14=2a+5d
(a+9d)/(a+12d)=1/3
3a+27d=a+12d
2a+15d=0
2a+5d=14
10d=-14
d=-1.4
2a-7=14
2a=21
a=10.5
T21=a+20d
= 10.5-20*1.4
= 10.5-28
=-17.5
I think I am wrong...check the steps and correct me in comments.
84=6[2a+5d]
14=2a+5d
(a+9d)/(a+12d)=1/3
3a+27d=a+12d
2a+15d=0
2a+5d=14
10d=-14
d=-1.4
2a-7=14
2a=21
a=10.5
T21=a+20d
= 10.5-20*1.4
= 10.5-28
=-17.5
I think I am wrong...check the steps and correct me in comments.
simran262:
ohh... ty soo much i will check and let u know about it
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