Question

If first, second and last terms of an AP are x,y and 2x respectively, then find the sum of AP.

Answer 1

hey dear



Here is your answer



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Let the AP series be x, x + d, x + 2d ..

x + ( n - 1 ) d




where

x is the first term of sequence


d is the arithmetic difference



n is the number of term



According to question




First term = x = a..... ( 1 )



second term = x + d = b.



d = b - a. ( 2)




last term x + ( n - 1 ) * d = 2x



substituting the value of d from the above equation 2 in the equation we get



( n - 1 ) ( y - x) = x



n - 1 = x / ( y - x)



n = y / ( y - x) .. ( 3)




Hence as we know sum terms of an AP





AP = n / 2 ( 2* first term + ( n -1 ) d)



Sum = Y / 2 ( y - x) 2x +x / ( y -x) ( y -x)




Sum = 3xy / 2 ( y - x)



hence your answer is 3xy / 2 ( y - x)




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hope it helps


thank you

Answer 2

Hey friend, Harish here.

Here is your answer:

Given:

First term = x 

Second term = y

Last term = 2x

To Find:

The sum of the AP.

Solution:

$\mathrm{Common\ difference\ (d) = Second\ term - First\ Term(a)} \\ \\ \to d = y - x \\ \\ \mathrm{We\ Know\ that\ :}\\ \\ \mathrm{n^{th} \ term = a+(n-1)d} \\ \\ \to 2x = x + (n-1)(y-x) \\ \\ \to x = (n-1)(y-x) \\ \\ \to (n-1) = \frac{x}{(y-x)} \\ \\ \to n = \frac{y}{(y-x)} \\ \\ \mathrm{Now, Sum \ of\ AP= \frac{n}{2}\times \Bigl( First\ term + Last\ term \Bigr ) } \\ \\ \to Sum = \frac{y}{2(y-x)} \times (x+2x) \\ \\ \to Sum = \frac{3xy}{2(y-x)}$
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Hope my answer is helpful to you.