Question

If the sum of the first n terms of an AP is \(nA+n^2B\) where A and B are constants, then the common difference of the AP is
a)A+B
b)A-B
c)2A
d)2B

Answer 1

It is D..

Answer 2

How do you get it?

Answer 3

See, if that is the sum of first n terms then:

Put n= 1, you will get the first term:

First term = A + B

Now putting n= 2 you will get the sum of First two terms:

Sum of First two terms : 2A + 4B
Now subtract both of them you will get the second term:

2A + 4B - A - B = A + 3B

This is the second term..

Now subtract second and first term you will get the common difference..

Answer 4

Sum of n-terms of an AP = \(\frac{n}{2}(2a + (n-1)d) \), a = first term, d = common difference
\[S = n(a-\frac{d}{2}) + n^{2}\frac{d}{2}\]
Comparing above equation with \(An + Bn^{2} \), we get
\[A = a - \frac{d}{2}, B = \frac{d}{2}\]
or \(d = 2B \)

Answer 5

Thanks @waterineyes and @FoolAroundMath

Answer 6

Welcome dear..

Answer 7

In any problem where sum is given to you in n form then find first and second term by this procedure and then find common difference...

Answer 8

yeah, i will be doing that now on wards! Thanks!!