Q. 263.6( 5 Votes )

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Answer :

Sum of n terms of an A.P., Sn =

Where a=First term, d=Common difference, n=No. in terms


For n=4, S4= (given)


2a + 3d=20


2a=20-3d ………….(1)


For n=14,S14= (given)


2a + 13d=40 ………..(2)


Putting the value of 2a in eq(2) from eq(1)


20-3d + 13d=40


20 + 10d=40


10d=20


d=2


Putting value of d in eq(1), we get


2a=20-3(2)=20-6=14


a=7


Sum of n terms, Sn =


Sn =


Sn =n[7 + n-1]


Sn =n(n + 6)=n2 + 6n


Or


The first 30 positive integers divisible by 6 are 6,12,18,24…..


We see that these are in A.P.


With a=6 and d=12-6=6


Sum of n terms, Sn =


S30=


S30=15(186)=2790


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