Question
Find the sum:$\text{7}+10\frac{1}{2}+14+\ .....\ + 84$
✓
Answer
Given,
$\text{A.P., }7+10\frac{1}{2}+14+\ .....\ + 84$
Here,
First term, $a = 7$
Difference, $\text{d}=\frac{21}{2}-7=\frac{7}{2}$
and Last term, $a_n = 84$
We know, $a_n = a + (n - 1)d$
$\Rightarrow\ 84= 7 + (\text{n}-1)\frac{7}{2}$
$\Rightarrow\ 84=7+\frac{7\text{n}}{2}-\frac{7}{2}$
$\Rightarrow\ 84=\frac{14+7\text{n}-7}{2}$
$\Rightarrow\ 84\times2=7+7\text{n}$
$\Rightarrow\ 168=7+7\text{n}$
$\Rightarrow\ 7\text{n}=168-7$
$\Rightarrow\ \text{n}=\frac{161}{7}=23$
We know, Sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\Big[2(7)+(23-1)\frac{7}{2}\Big]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\Big[14+22\times\frac{7}{2}\Big]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}[14+77]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\times91=\frac{2093}{2}$
Hecne, Sum of given A.P. is $\frac{2093}{2}$.
$\text{A.P., }7+10\frac{1}{2}+14+\ .....\ + 84$
Here,
First term, $a = 7$
Difference, $\text{d}=\frac{21}{2}-7=\frac{7}{2}$
and Last term, $a_n = 84$
We know, $a_n = a + (n - 1)d$
$\Rightarrow\ 84= 7 + (\text{n}-1)\frac{7}{2}$
$\Rightarrow\ 84=7+\frac{7\text{n}}{2}-\frac{7}{2}$
$\Rightarrow\ 84=\frac{14+7\text{n}-7}{2}$
$\Rightarrow\ 84\times2=7+7\text{n}$
$\Rightarrow\ 168=7+7\text{n}$
$\Rightarrow\ 7\text{n}=168-7$
$\Rightarrow\ \text{n}=\frac{161}{7}=23$
We know, Sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\Big[2(7)+(23-1)\frac{7}{2}\Big]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\Big[14+22\times\frac{7}{2}\Big]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}[14+77]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\times91=\frac{2093}{2}$
Hecne, Sum of given A.P. is $\frac{2093}{2}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the following systems of equations:
$\frac{10}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=4,$
$\frac{15}{\text{x}+\text{y}}-\frac{5}{\text{x}-\text{y}}=-2.$
→$\frac{10}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=4,$
$\frac{15}{\text{x}+\text{y}}-\frac{5}{\text{x}-\text{y}}=-2.$
ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersect AE at F, prove that AF × FB = EF × FD.
In the given figure, $\angle\text{ABC}=90^\circ$ and $\text{BD}\perp\text{AC}.$ If AB = 5.7cm, BD = 3.8cm, and CD = 5.4cm find BC.
→Mr. Ramesh Sawant invested ₹ 2,50,295 in shares of face value ₹ 100 each at ₹ 250 market value. He gave brokerage of 0.1 % and GST of 18 % on brokerage then how many shares are purchased by him?
Solve the following quadratic equation.
$m^2+5 m+5=0$
→$m^2+5 m+5=0$
In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.
Show that $A (1, 2) , (1, 6), C (1+2 \sqrt{3}, 4)$ are vertices of an equilateral triangle.
→Fill in the blanks given in the contract note of sale-purchase of shares.
(B - buy S - sell)
(B - buy S - sell)
| No. of shares | MV of shares | Total value | Brokerage 0.2% | 9% CGST on brokerage | 9% SGST on brokerage | Total value of shares |
| 100 B | Rs. 45 | |||||
| 75 S | Rs. 200 |
A solid is in the form of a right circular done mounted on a hemisphere. The radius of the hemisphere is $2.1\ cm$ and the height of the cone is $4\ cm.$ The solid is placed in a cylinrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the cylinder is 5cm and its height is $9.8\ cm$, find the volume of the water left in the tub.
→Solve the following system of equations by the method of cross-multiplication:
$\text{ax}+\text{by}=\frac{\text{a}+\text{b}}{2}$
$3\text{x}+5\text{y}=4$
→$\text{ax}+\text{by}=\frac{\text{a}+\text{b}}{2}$
$3\text{x}+5\text{y}=4$