Question

The first and third terms of an arithmetic progression are equal,respectively, to the first and third term of a geometric progression, and the second term of the AP exceeds the 2nd term of the GP by 0.25. calculate the sum of the first five terms of the arithmetic progression if its first term is equal to 2.?

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Answer 1

Answer:  The required sum of the first five terms of the A.P. is $\dfrac{5}{2}$ or $\dfrac{45}{2}.$

Step-by-step explanation:  Given that the first and third terms of an A.P. are respectively equal to the first and third terms of G.P. and the second  term of the AP exceeds the 2nd term of the GP by 0.25.

Also, the first term of both the progressions is 2.

We are to calculate the sum of first five terms of the A.P.

Let a be the first term of both A.P. and G.P., d and r be the common difference of the A.P. and common ratio of the G.P. respectively.

Then, according to the given information, we have

$\textup{third term of A.P.}=\textup{third term of G.P.}\\\\\Rightarrow a+(3-1)d=ar^{3-1}\\\\\Rightarrow a+2d=ar^2\\\\\Rightarrow 2+2d=2r^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since first term a is 2}]\\\\\Rightarrow r^2=1+d\\\\\Rightarrow d=r^2-1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$a+(2-1)d-ar^{2-1}=0.25\\\\\Rightarrow a+d-ar=\dfrac{1}{4}\\\\\\\Rightarrow 2+d-2r=\dfrac{1}{4}\\\\\\\Rightarrow 2+r^2-1-2r=\dfrac{1}{4}~~~~~~~~~~~~~~~~~~~~[\textup{Using equation (i)}]\\\\\\\Rightarrow r^2-2r+1=\dfrac{1}{4}\\\\\Rightarrow 4r^2-8r+3=0\\\\\Rightarrow 4r^2-6r-2r+3=0\\\\\Rightarrow 2r(2r-3)-1(2r-3)=0\\\\\Rightarrow (2r-1)(2r-3)=0\\\\\Rightarrow 2r-1=0,~~~~~2r-3=0\\\\\Rightarrow r=\dfrac{1}{2},\dfrac{3}{2}.$

If $r=\dfrac{1}{2},$ then from equation (i), we have

$d=\dfrac{1}{4}-1=-\dfrac{3}{4}.$

If $r=\dfrac{3}{2},$ then from equation (i), we have

$d=\dfrac{9}{4}-1=\dfrac{5}{4}.$

So, the sum of first five term of the A.P. will be

$S_5=\dfrac{5}{2}(2\times 2+(5-1)\times(-\dfrac{3}{4}))=\dfrac{5}{2}(4-3)=\dfrac{5}{2}$

or

$S_5=\dfrac{5}{2}(2\times 2+(5-1)\times(\dfrac{5}{4}))=\dfrac{5}{2}(4+5)=\dfrac{45}{2}$

Thus, the required sum of the first five terms of the A.P. is $\dfrac{5}{2}$ or $\dfrac{45}{2}.$