any answer would be appreciatedSolutionb Sum of n terms of A

any answer would be appreciated

Solution

b)

Sum of n terms of AP= n(+1)/2

(2i -1) = 2i - 1 = 2*{n*(n+1)/2} - n = n^2 + n - n = n^2

e)

Expading the series we get {a, ar, ar22, ar33, ar44, ...........} i.e GP with first term ‘a’ and common ratio r. Then,

Now, the nth terms of the given GP = a r
n1.

Sn = a + ar + ar2 + ar3 + ar4 + ........ + arn2 + arn1......... (i)

Multiplying both sides by r, we get,

rSn = ar + ar2 + ar3 + ar4 + ar5+ ................ + arn1 + arn ............ (ii)

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On subtracting (ii) from (i), we get

Sn - rSn = a - arn

Sn(1 - r) = a(1 - rn)

Sn = a(1rn)(1r)

or Sn= a(rn1)(r1)

f) Let x be any integer. So next six integers will be x+1,x+2,x+3,x+4,x+5,x+6.

Sum of integers= x+ (x+1)+(x+2)+(x+3)+(x+4)+x+5)+(x+6) =7x+21= 7(x+3)

We can easily see irrespective of the number the sum of seven consiquitive number have 7 as one of the factor. So divide 7(x+3) by 7, the sum is always divisible by 7.