Question
Let there be an A.P. with first term ' $a$ ', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find. d , if $a =3, n =8$ and $S _{ n }=192$.
✓
Answer
Here, we have an A.P. whose first term (a), Sum of first n terms ($S_n$) and the number of terms (n) are given. We need to find common difference (d).
Here,
First term (a) = 3
Sum of n terms ($S_n$) = 192
Number of terms (n) = 8
So here we will find the value of n using the formula, $a_n = a + (n - 1)d$
So. to find the common difference of this A.P., we use the following formula for the sum of n terms of an A.P.
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 8, we get,
$\text{S}_8=\frac{8}{2}[2(3)+(8-1)(\text{d})]$
$192 = 4[6 + (7)(d)]$
$192 = 24 + 28d$
$28d = 192 - 24$
Furhter solving for d,
$\text{d}=\frac{168}{28}$
$d = 6$
Therefore, the common difference of the given A.P. is d = 6.
Here,
First term (a) = 3
Sum of n terms ($S_n$) = 192
Number of terms (n) = 8
So here we will find the value of n using the formula, $a_n = a + (n - 1)d$
So. to find the common difference of this A.P., we use the following formula for the sum of n terms of an A.P.
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 8, we get,
$\text{S}_8=\frac{8}{2}[2(3)+(8-1)(\text{d})]$
$192 = 4[6 + (7)(d)]$
$192 = 24 + 28d$
$28d = 192 - 24$
Furhter solving for d,
$\text{d}=\frac{168}{28}$
$d = 6$
Therefore, the common difference of the given A.P. is d = 6.
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
An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45cm and 25cm, the total vertical height of the bucket is 40cm and that of the cylindrical base is 6cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.
In the given figure, $\angle1=\angle2$ and $\frac{\text{AC}}{\text{BD}}=\frac{\text{CB}}{\text{CE}}$
Prove that $\triangle\text{ACB}\sim\triangle\text{DCE}.$
→Prove that $\triangle\text{ACB}\sim\triangle\text{DCE}.$
Draw a circle of radius 4cm. Construct a pair of tangents to it, the angle between which is 60º. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.
Prove thales theorem.
Solve the following system of equations graphically:
2x + 3y - 4 = 0,
3x - y + 5 = 0
2x + 3y - 4 = 0,
3x - y + 5 = 0
Prove the following identities:
$\frac{\tan\theta}{(1-\cot\theta)}+\frac{\cot\theta}{(1-\tan\theta)}=(1+\sec\theta\text{ cosec }\theta)$
→$\frac{\tan\theta}{(1-\cot\theta)}+\frac{\cot\theta}{(1-\tan\theta)}=(1+\sec\theta\text{ cosec }\theta)$
In the given figure, each one of PA , QB and RC is perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that $\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{\text{z}}.$
→Compute the mode from the following series:
|
Size
|
45-55
|
55-65
|
65-75
|
75-85
|
85-95
|
95-105
|
105-115
|
|
Frequency
|
7
|
12
|
17
|
30
|
32
|
6
|
10
|
Solve the following quadratic equations by factorization:
$ax^2 + (4a^2 - 3b)x - 12ab = 0$
→$ax^2 + (4a^2 - 3b)x - 12ab = 0$
A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs. 216 and one full and one half reserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the reservation charge?