Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
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Let 1st term of AP be a
Its common difference
d = 3rd term - 2nd term = 18-14=4
So a+d =14=> a=14-4=10
Sum of first 51 terms =51/2(2*10+(51-1)*4)
= 51×110=5610
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given,
`a_2=14`
`a_3=18`
d=`a_3-a_2=4`
`a_1=a_2-d`
`a_1`=14-4
a=10
`S_n`=n/2(2a+(n-1)d)
`S_n`=51/2(2*10+(51-1)4)
`S_n`=5610
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