an ap consists of 50 terms of which 3rd term is 12 and last term is 106 find the sum of first 10 numbers
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Answer:
a3=12
n=3
An=a+(n-1)d
=a+(3-1)d. _(1)
An=106
106=a+(50-1)d
106=a+49d. _(2)
from (1)&(2)
106=a+49d
12=a+2d
-. -. -
94=47d
d=94/47
d=2
12=a+2d
12=a+2×2
12=a+4
a=12-4
a=8
Sn=n/2[2a+(n-1)d]
= 10/2[2(8)+(10-1)2]
= 5(16+(9)2)
=5(16+18)
=5(34)
Sn =170
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