Question

1. If the sum of the first n terms of a sequence is
of the form An2 + Bn, where A, B are constants,
independent of n, show that the sequence is an
A.P. Is the converse true? Justify your answer.​

Answer 1

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☆GIVEN

Sum of first n term is :-An²+Bn where A and b are constant

So, we have 2 show the sequence is ap

now,let take n = 1

so value come

An²+Bn

=>A+B (1)

again let n=2

=> 4a+2b (2)

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subtracting eq 1 from 2..

2a+2b-(a+b)

=4a+2b-a-b

=3a+b

So this is first term (3a+b)

similarly we find second term..

when n=3

=>9a+3b (3)

Now agin subtracting equation 2 from 3

we get. term 2

9a+3b-4a-2b

=5a-b

Now cd(T2-T1)

(5a-b-3a-b)

d=2a-2b

Let check this is ap or not

$\fbox\color{brown}{</em></strong><strong><em>T</em></strong><strong><em>n=</em></strong><strong><em>a+(n-1)d}$

a=3a+b

d=2a-2b

n=2

so putting these value in tn formula

3a+b+2a-2b

=5a-b. (4)

Here the the equation (4) is equal value to the term 2

Hence, This is an ap

yes it is converse

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