Arithmetic Progression: Sum of n Terms

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

Sum of first $$n$$ terms of an A.P. is given by  $$S_n = \dfrac{n}{2} \big[2a + (n-1)d \big] $$

Now, for the first 4 terms: $$ S_4 = \dfrac{4}{2}[2a + 3d] = 2(2a + 3d) = 6 \Rightarrow 2a + 3d = 3 \quad ...(1) $$

And, for first 6 terms: $$ S_6 = \dfrac{6}{2}[2a + 5d] = 3(2a + 5d) = 4 \Rightarrow 2a + 5d = \dfrac{4}{3} \quad ...(2) $$

Subtracting (1) from (2),
$$ (2a + 5d) - (2a + 3d) = \dfrac{4}{3} - 3 $$

$$ \Rightarrow 2d = \dfrac{4}{3} - \dfrac{9}{3} = -\dfrac{5}{3} \Rightarrow d = -\dfrac{5}{6} $$

Substituting this into (1):

$$ 2a + 3\left(-\dfrac{5}{6}\right) = 3 $$

$$ \Rightarrow 2a - \dfrac{5}{2} = 3 $$

$$ \Rightarrow a = \dfrac{11}{4} $$

Now, $$ S_{12} = \dfrac{12}{2}[2a + 11d] = 6[2a + 11d] $$

$$=6\ \cdot\left[\dfrac{11}{2}-\dfrac{55}{6}\right]=6\ \cdot\dfrac{33-55}{6}=-22$$

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.