Question

find (4-1/n )+(7-2/n)+(10-3/n)............upto n terms class 10

Answer 1

(4-1/n)+(7-2/n)+(10-3/n)+........n terms

(4+7+10+..... n term)-1/n{1+2+3+4.....n term)

here you see two series .
for first series ,
==============
a=4
d=3
Sn=n/2{2a+(n-1)d} {AP formula
now,
Sn=n/2{2.4+(n-1)3}=n/2(3n+5)======(1)

for second series
(1+2+3+4+.....n terms )
this is natural number so,
Sn=n(n+1)/2 ============(2)
now put this value above

n/2(3n+5)-(n+1)/2
1/2{3n^2+5n-n-1}=1/2{3n^2+4n-1}

Answer 2

Please find below the solution to the asked query:

We have Arithmetic Progressions , As : ( 4 − 1n ) + ( 7 − 2n ) + ( 10 − 3n ) + .... n− terms

Here common difference =  d = ( 7 − 2n ) − ( 4 − 1n ) = ( 10 − 3n ) − ( 7 − 2n ) =( 3 −1n)

We know , formula of sum of n terms of arithmetic progression is 

S = n2[ 2a1 + ( n - 1 ) d ] , So

Sum of n terms of given A.P.:

⇒n2[2 ( 4 − 1n)+ ( n − 1 )( 3 −1n)]⇒n2[(8− 2n)+ (  3 n − 1 − 3 + 3n) ]⇒n2[ 4+ 1n +3 n ]