Find the sum of 50 terms of an AP whose third term is 5 and the seventh term is 9.
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sanchi2003:
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a +2d=5
a+6d=9
Subtract the 2 equation
-4d=-4
D=1
So a is 5-2= 3
( I have substituted the value of different in equation 1)
Now sum of 50 terms
S=n/2(2a+(n-1)d)
= 25( 6+24X1)
= 25X30
= 750
a+6d=9
Subtract the 2 equation
-4d=-4
D=1
So a is 5-2= 3
( I have substituted the value of different in equation 1)
Now sum of 50 terms
S=n/2(2a+(n-1)d)
= 25( 6+24X1)
= 25X30
= 750
hope it helps
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