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Question

Find the sum of first n terms of the series 3+7+13+21+31+....

Or

If a b and c are in GP and x,y are the arithmetic means of a,b and b,c respectively. prove that ax+cy=2 and 1x+1y=2b.

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Solution

We have,the given series 3+7+13+21+31+...

The difference between the two successive terms are

73=4,13=6,2113=8

3121=10,are in AP

Let an be the nth terms and Sn denotes the sum of n terms.

Then,Sn=3+7+13+21+31+...+an1+an ..(i)

Also,Sn=3+7+13+21+...+an2+an1+an ..(ii)

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

0=3+4+6+8+10+...+(anan1)an

an=3+4+6+8+10+...+(anan1)

an=3+n12[Sn=n2[2a+(n1)d]]

an=3+n12(8+2n4)

an=n2+n+1

Sn=an=(n2+n+1)

=n(n+1)(2n+1)6+n(n+1)2+n

=n[(n+1)(2n+1)6+(n+1)2+1]

=n[(2n+3n+1)+3n+3+66]

=n6[2n2+6n+10]=n3(n2+3n+5)

it is given that a,b and c are in GP.

b2=ac

Also given x is the AM of a and b.

x=a+b2

and y is the AM of b and c.

y=b+c2

Now ,ax+cy=aa+b2+cb+c2

=2aa+b+2cb+c

=2a(b+c)+2c(a+b)(a+b)(b+c)

=2(ab+ac+ac+bc)ab+ac+b2+bc

=2(ab+2ac+b)ab+ac+a+bc [b2=ac]

=2(ab+2ac+bc)ab+2ac+bc=2

and =1x+=1y=1=a+b2+1=b+2

=2a+b+2b+c=2(b+c+a+b)(a+b)(b+c)

=2(a+c+2b)ab+b2+ac+bc

=2(a+c+2b)ab+b2+b2+bc [ac=b2]

=2(a+c+2b)b(a+2b+c)=2b

=1x+1y=2b Hence proved.


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