Arithmetic Progression

The arithmetic progression has a total of $$2K$$ terms, where, $$K$$ is natural number.

Let the first term of the A.P be $$a$$ and the common difference be $$d$$. 

Last term of the A.P = $$a+\left(2K-1\right)d$$

Difference between the last term and the first term of the A.P = $$a+\left(2K-1\right)d\ -a=\left(2K-1\right)d$$

Difference = 27 (given) 

$$\left(2K-1\right)d=27$$                                                            $$\longrightarrow\ i$$

Now, it is given that the sum of all the odd and even terms are 40 and 55 respectively. 

Odd terms : 1st term, 3rd term, 5th term, ....., $$(2K-1)$$th term

Odd terms : $$a,\ (a+2d),\ (a+4d),\ .....,\ (a+(2K-2)d)$$

Total number of odd terms = $$K$$

Sum of all the odd terms = $$\dfrac{K}{2}\times\left(a+\left(a+\left(2K-2\right)d\right)\right)=40$$

$$\dfrac{K}{2}\times\left(2a+\left(2K-1\right)d-d\right)=40$$

$$\dfrac{K}{2}\times\left(2a+27-d\right)=40$$                                 $$\longrightarrow\ ii$$

Even terms : 2nd term, 4th term, 6th term, ....., $$(2K)$$th term

Even terms : $$(a+d),\ (a+3d),\ (a+5d),\ .....,\ (a+(2K-1)d)$$

Total number of even terms = $$K$$ 

Sum of all the even terms = $$\dfrac{K}{2}\times\left(\left(a+d\right)+\left(a+\left(2K-1\right)d\right)\right)=55$$

$$\dfrac{K}{2}\times\left(2a+27+d\right)=55$$                                 $$\longrightarrow\ iii$$

Divide equation $$iii$$ by equation $$ii$$, 

$$\dfrac{\left(2a+27\right)+d}{\left(2a+27\right)-d}=\dfrac{55}{40}=\dfrac{11}{8}$$

$$8\left(2a+27\right)+8d=11\left(2a+27\right)-11d$$

$$3\left(2a+27\right)=19d$$

$$2a+27=\dfrac{19d}{3}$$                                                               $$\longrightarrow\ iv$$

Insert the value of $$(2a+27)$$ in equation $$ii$$ from equation $$iv$$,

$$\dfrac{K}{2}\times\left(\dfrac{19d}{3}-d\right)=40$$

$$\dfrac{K}{2}\times\left(\dfrac{16d}{3}\right)=40$$

$$Kd=15$$                                                                                    $$\longrightarrow\ v$$

Insert the value of $$d$$ from equation $$v$$ to equation $$i$$ to find the value of $$K$$. 

$$\left(2K-1\right)\times\dfrac{15}{K}=27$$

$$\left(2K-1\right)\times5=9K$$

$$10K-5=9K$$

$$K=5$$

Hence, the value of $$K$$ is 5. 

$$\therefore\ $$ The required answer is B. 

Let the roots of $$3x^{2} - px + q = 0$$ be $$\alpha$$ and $$\beta$$. 

It is given that $$\alpha$$ and $$\beta$$ are the 10th and 11th term of an A.P with a common difference of $$\dfrac{3}{2}$$.

10th term = $$a+9d$$ = $$a+\left(9\times\dfrac{3}{2}\right)$$ = $$a+\dfrac{27}{2}$$

11th term = $$a+10d$$ = $$a+\left(10\times\dfrac{3}{2}\right)$$ = $$a+15$$

Now, it is given that the sum of the first 11 terms of the A.P is 88. 

$$\dfrac{11}{2}\times\left(a+\left(a+15\right)\right)=88$$

$$2a+15=16$$

$$a=\ \dfrac{1}{2}$$

10th term = $$\dfrac{1}{2}+\dfrac{27}{2}=14$$

11th term = $$\dfrac{1}{2}+15=\dfrac{31}{2}$$

So, $$\alpha=14$$ and $$\beta=\dfrac{31}{2}$$

Now, in $$3x^{2} - px + q = 0$$, 

$$\alpha+\beta=\dfrac{p}{3}$$

$$14+\dfrac{31}{2}=\dfrac{p}{3}$$

$$\dfrac{28+31}{2}=\dfrac{p}{3}$$

$$p=\dfrac{59\times3}{2}$$

$$\alpha\times\beta=\dfrac{q}{3}$$

$$14\times\dfrac{31}{2}=\dfrac{q}{3}$$

$$q=7\times31\times3$$

$$q=651$$

Now, $$q-2p$$ = $$651-\left(2\times\dfrac{59\times3}{2}\right)$$

$$q-2p$$ = $$651-177=474$$

Hence, the value of $$(q-2p)$$ is 474. 

$$\therefore\ $$ The required answer is 474.

$$S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\cdots$$ upto $$n$$ terms.

$$S_{2025}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+.....\ 2025\ terms$$

$$S_{2025}=\dfrac{2-1}{1\times2}+\dfrac{3-2}{3\times2}+\dfrac{4-3}{4\times3}+....\ +\dfrac{2026-2025}{2025\times2026}$$

$$S_{2025}=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+.....+\left(\dfrac{1}{2025}-\dfrac{1}{2026}\right)$$

$$S_{2025}=1-\dfrac{1}{2026}$$

$$2026\times S_{2025}=2026\times\frac{2025}{2026}=2025$$

The sum of the first six terms of an A.P. with first term $$-p$$ and common difference $$p$$ is $$\sqrt{2026\, S_{2025}}$$

$$-p+0+p+2p+3p+4p=\sqrt{2025}$$

$$9p=45$$

$$p=5$$

$$a_{20}=a+19d=(-p)+19p=18p$$

$$a_{15}=a+14d=(-p)+14p=13p$$

Thus, $$a_{20}-a_{15}=18p-13p=5p=5\times5=25$$