Sum of the first n terms of an A.P having first term a and last term l is
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let common difference be d
then second term will be a+(2-1)d=a+d
then third term will be
a+d+d=a+(3-1)d=a+2d
then fourth term will be
a+2d+d=a+(4-1)d=a+3d
then sum of first two terms is 2a+d
then sum of first three terms is 3a+3d
then sum of first four terms is 4a+6d
then we can write the above sum as
2a+d= 2a+2(2-1)d/2
& 3a+3d=3a+3(3-1)d/2
& 4a+6d=4a+4(4-1)d/2
then nth term will be a+(n-1)d=l
and sum of first nth term will be
na+n(n-1)d/2
=n/2(2a+(n-1)d)
=n/2(a+a+(n-1)d)
=n/2(a+l).......(ans)
then second term will be a+(2-1)d=a+d
then third term will be
a+d+d=a+(3-1)d=a+2d
then fourth term will be
a+2d+d=a+(4-1)d=a+3d
then sum of first two terms is 2a+d
then sum of first three terms is 3a+3d
then sum of first four terms is 4a+6d
then we can write the above sum as
2a+d= 2a+2(2-1)d/2
& 3a+3d=3a+3(3-1)d/2
& 4a+6d=4a+4(4-1)d/2
then nth term will be a+(n-1)d=l
and sum of first nth term will be
na+n(n-1)d/2
=n/2(2a+(n-1)d)
=n/2(a+a+(n-1)d)
=n/2(a+l).......(ans)
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