An ap consists of 50 terms of which 3rd term is 12 and last term is 106. Find the 25th term
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Let common difference be d & the first term be a
3rd TERM=a+(3-1)d=a+2d=12--eq.1
50th TERM=a+(50-1)d=a+49d=106 --eq2
Subtracting eq 1 from 2 we get:
a+49d=106
a+2d=12
------------------
47d=94
d=94/47=2
a=12-2d=12-4=8
25th TERM=a+24d=8+48=56
Hope it helps
3rd TERM=a+(3-1)d=a+2d=12--eq.1
50th TERM=a+(50-1)d=a+49d=106 --eq2
Subtracting eq 1 from 2 we get:
a+49d=106
a+2d=12
------------------
47d=94
d=94/47=2
a=12-2d=12-4=8
25th TERM=a+24d=8+48=56
Hope it helps
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