Find the sum of the two middle most terms of the A.P. \[-\mathbf{4}/\mathbf{3},\text{ }-\mathbf{1},\text{ }-\mathbf{2}/\mathbf{3},\text{ }\ldots ,4\frac{1}{3}\]

Find the sum of the two middle most terms of the A.P.

$-\mathbf{4}/\mathbf{3},\text{ }-\mathbf{1},\text{ }-\mathbf{2}/\mathbf{3},\text{ }\ldots ,4\frac{1}{3}$

Find the sum of the two middle most terms of the A.P.

$-\mathbf{4}/\mathbf{3},\text{ }-\mathbf{1},\text{ }-\mathbf{2}/\mathbf{3},\text{ }\ldots ,4\frac{1}{3}$

From the question,

Last term (n^th) =

$4\frac{1}{3}$

$13/3$

First term a =

$-4/3$

Then, difference

$\begin{array}{*{35}{l}} d\text{ }=\text{ }-1\text{ }\text{ }\left( -4/3 \right)\text{ }=\text{ }-1\text{ }+\text{ }4/3\text{ }=\text{ }\left( -3\text{ }+\text{ }4 \right)/3\text{ }=\text{ }1/3 \\ =\text{ }-2/3\text{ }\text{ }\left( -1 \right)\text{ }=\text{ }-2/3\text{ }+\text{ }1\text{ }=\text{ }\left( -2\text{ }+\text{ }3 \right)/3\text{ }=\text{ }1/3 \\ \end{array}$

Therefore, common difference d =

$1/3$

We know that,

$\begin{array}{*{35}{l}} {{T}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ So,\text{ }13/3\text{ }=\text{ }-4/3\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\left( 1/3 \right) \\ 13/3\text{ }+\text{ }4/3\text{ }=\text{ }1/3n\text{ }\text{ }1/3 \\ 13/3\text{ }+\text{ }4/3\text{ }+\text{ }1/3\text{ }=\text{ }1/3n \\ \left( 13\text{ }+\text{ }4\text{ }+\text{ }1 \right)/3\text{ }=\text{ }1/3n \\ 18/3\text{ }=\text{ }1/3n \\ 6\text{ }=\text{ }1/3n \\ n\text{ }=\text{ }6\text{ }\times \text{ }3 \\ n\text{ }=\text{ }18 \\ \end{array}$

So, middle term is

$18/2$

and

$\left( 18/2 \right)\text{ }+\text{ }1\text{ }=\text{ }{{9}^{th}}~$

and

${{10}^{th}}$

term

Then,

$\begin{array}{*{35}{l}} ~{{a}_{9}}~+\text{ }{{a}_{10}}~=\text{ }a\text{ }+\text{ }8d\text{ }+\text{ }a\text{ }+\text{ }9d \\ =\text{ }2a\text{ }+\text{ }17d \\ \end{array}$

Now substitute the value of ‘a’ and ‘d’.

$\begin{array}{*{35}{l}} =\text{ }2\left( -4/3 \right)\text{ }+\text{ }17\left( 1/3 \right) \\ =\text{ }-8/3\text{ }+\text{ }17/3 \\ =\text{ }\left( -8\text{ }+\text{ }17 \right)/3 \\ =\text{ }9/3 \\ =\text{ }3 \\ \end{array}$

Therefore, the sum of the two middle most terms of the A.P is

$3$

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

Two cells of emf $\varepsilon_{1}$ and $\varepsilon_{2}$ , internal resistance $r_{1}$ and $r_{2}$ , connected in parallel. The equivalent emf of the combination is- (A) $\frac{\varepsilon_{1} r_{1}+\varepsilon_{1} r_{1}}{r_{1}+r_{2}}$ (B) $\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$ (C) $\sqrt{\varepsilon_{1} \times \varepsilon_{2}}$ (D) $\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

If the points $A(a, 0), B(0, b)$ and $C(1,1)$ are collinear, prove that $\frac{1}{a}+\frac{1}{b}=1$

If $A(-2,0), B(0,4)$ and $C(0, k)$ be three points such that area of a $A B C$ is 4 sq units, find the value of $k$ .

Find the value of $k$ for which the area of a ABC having vertices $A(2,-6), B(5,4)$ and $C(k, 4)$ is 35 sq units.

Find the value of $k$ for which the points $A(1,-1), B(2, k)$ and $C(4,5)$ are collinear.

Find the value of $k$ for which the points $P(5,5), Q(k, 1)$ and $R(11,7)$ are collinear.

Find the value of $k$ for which the points $A(3,-2), B(k, 2)$ and $C(8,8)$ are collinear.

Use determinants to show that the following points are collinear. $P(-2,5), Q(-6,-7)$ and $R(-5,-4)$

Sign up now and study all subjects for free

Class

Syllabus

Question Papers

Social Media

Apps