For an AP,
$a_n$ = a+(n-1)d
= 28+(7-1)(-4)
= 28+6(-4)
= 28-24
$a_n$=4
a = 10, d = 10
$a_1$ = a = 10
$a_2 = a_1$+d = 10+10 = 20
$a_3 = a_2$+d = 20+10 = 30
$a_4 = a_3$+d = 30+10 = 40
First term, a = 3
Common difference, d = Second term – First term
1 – 3 = -2
d = -2
Given, 3, 8, 13, 18, … is the AP.
First term, a = 3
Common difference, $d = a_2 − a_1$ = 8 − 3 = 5
Let the nth term of given A.P. be 78. Now as we know,
$a_n = a+(n−1)d$
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
First term = -3 and second term = 4
a = -3
d = 4-a = 4-(-3) = 7
$a_n$ = a+(n-1)d
$a_{21}$=a+(21-1)d
=-3+(20)7
=-3+140
=137
The first five multiples of 3 is 3, 6, 9, 12 and 15
a=3 and d=3
n=5
Sum,$S_n = \dfrac{n}{2} [2a +(n – 1)d]$
$S_5 = \dfrac{5}{2}$[2(3)+(5-1)3]
$=\dfrac{5}{2}$[6+12]
$=\dfrac{5}{2}$[18]
=5 x 9
= 45
Given AP: 5, 8, 11, 14,….
First term = a = 5
Common difference = d = 8 – 5 = 3
$n^{th}$ term of an AP = $a_n$ = a + (n – 1)d
Now, $10^{th}$ term = $a_{10}$ = a + (10 – 1)d
= 5 + 9(3)
= 5 + 27
= 32
–10, –6, –2, 2,…
Let $a_1 = -10, a_2 = -6, a_3 = -3, a_4 = 2$
$a_2 – a_1$ = -6 – (-10) = 4
$a_3 – a_2$ = -2 – (-6) = 4
$a_4 – a_3$ = 2 – (-2) = 4
The given list of numbers is an AP with d = 4.
The famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.
Given AP: 10, 6, 2,…
Here, a = 10, d = -4
Sum of first n terms = $S_n = \dfrac{n}{2} [2a +(n – 1)d]$
The sum of first 16 terms = $S_{16} = (\dfrac{16}{2})$[2(10) + (16 – 1)(-4)]
= 8[20 + 15(-4)]
= 8(20 – 60)
= 8(-40)
= -320