Find the sum of 1st 51 terms of an AP whose second and third terms are 14 and 18 resp.
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Second term = a+d= 14
Third term = a+2d=18
Subtract these
a+2d-(a+d)= 18-14
So d = 4
as a+d=14
a= 10
Sum of n terms = n/2(2a+(n-1)d)
n=51 , a= 10 , d=4
Sum = 51/2(20+50*4)
=5610
If I deserve brainliest , please give it
Third term = a+2d=18
Subtract these
a+2d-(a+d)= 18-14
So d = 4
as a+d=14
a= 10
Sum of n terms = n/2(2a+(n-1)d)
n=51 , a= 10 , d=4
Sum = 51/2(20+50*4)
=5610
If I deserve brainliest , please give it
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4
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