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Sum of the first n terms:
For a sequence of real numbers $$x_{1},x_{2},...x_{n}$$, If $$x_{1}-x_{2}+x_{3}-....+(-1)^{n+1}x_{n}=n^{2}+2n$$ for all natural numbers n, then the sum $$x_{49}+x_{50}$$ equals
Now as per the given series :
we get $$x_1=1+2\ =3$$
Now $$x_1-x_2=\ 8$$
so$$x_2=-5$$
Now $$x_1-x_2+x_3\ =\ 15$$
so $$x_3\ =7$$
so we get $$x_n\ =\left(-1\right)^{n+1}\left(2n+1\right)$$
so $$x_{49}\ =\ 99$$ and $$x_{50}\ =-101$$
Therefore $$x_{49\ }+x_{50}\ =-2$$
Consider a sequence of real numbers, $$x_{1},x_{2},x_{3},...$$ such that $$x_{n+1}=x_{n}+n-1$$ for all $$n\geq1$$. If $$x_{1}=-1$$ then $$x_{100}$$ is equal to
Given $$x_{n+1}\ =\ x_n\ +\ n\ -1$$ and x1 = -1.
Considering
x1 = -1. (1)
x2 = x1+1-1 = x1 + 0 (2)
x3 = x2 + 2 - 1 =x2 + 1 (3)
x4 = x3 + 3 - 1 = x3 + 2 (4)
x100 = x99 + 98 (100)
Adding the LHS and RHS for the hundred equations we have:
(x1+x2+......................x100) = (-1+0+.........98) + (x1+x2+...............x99)
Subtracting this we have :
(x1+...........x100) - (x1+............. x 99) = $$\frac{\left(98\cdot99\right)}{2}$$ - 1.
x100 = 4851 - 1 = 4850
Alternatively
$$x_1=-1$$
$$x_2=x_1+1-1=x_1=-1$$
$$x_3=x_2+2-1=x_2+1=-1+1=0$$
$$x_4=x_3+3-1=x_3+2=0+2=2$$
$$x_5=x_4+4-1=x_4+3=2+3=5$$
......
If we observe the series, it is a series that has a difference between the consecutive terms in an AP.
Such series are represented as $$t\left(n\right)=a+bn+cn^2$$
We need to find t(100).
t(1) = -1
a + b + c = -1
t(2) = -1
a + 2b + 4c = -1
t(3) = 0
a + 3b + 9c = 0
Solving we get,
b + 3c = 0
b + 5c = 1
c = 0.5
b = -1.5
a = 0
Now,
$$t\left(100\right)=\left(-1.5\right)100+\left(0.5\right)100^2=-150+5000=4850$$
For any natural number n, suppose the sum of the first n terms of an arithmetic progression is $$(n + 2n^2)$$. If the $$n^{th}$$ term of the progression is divisible by 9, then the smallest possible value of n is
It is given,
$$S_n=2n^2+n$$
$$S_{n-1}=2\left(n-1\right)^2+\left(n-1\right)$$
$$S_{n-1}=2n^2-3n+1$$
$$T_n=S_n-S_{n-1}=2n^2+n-2n^2+3n-1=4n-1$$
$$T_n=4n-1$$
The terms are 3, 7, 11, 15, 19, 23, 27,......
27 is the first term in the series divisible by 9.
27 is the 7th term.
Therefore, the least possible value of n is 7.
The answer is option C.
Consider the arithmetic progression 3, 7, 11, ... and let $$A_n$$ denote the sum of the first n terms of this progression. Then the value of $$\frac{1}{25} \sum_{n=1}^{25} A_{n}$$ is
Sum of n terms in an A.P = $$\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)$$
$$A_n=\dfrac{n}{2}\left(6+\left(n-1\right)4\right)=n\left(2n+1\right)$$
$$\Sigma\ A_n=\Sigma\ n\left(2n+1\right)=2\Sigma\ n^2+\Sigma\ n=\ \dfrac{\ 2n\left(n+1\right)\left(2n+1\right)}{6}+\ \dfrac{\ n\left(n+1\right)}{2}$$
Substituting n = 25, we get
$$\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\dfrac{1}{25}\left(\ \dfrac{\ 2\left(25\right)\left(25+1\right)\left(50+1\right)}{6}+\ \dfrac{\ 25\left(25+1\right)}{2}\right)$$
$$\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\ 26\left(17\right)+13$$ = 455
The answer is option A.
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