Arithmetic
Progressions
3) In an AP.
(viii) Given a12 = 37, d = 3, find a and S12.
Sol:
a12
= a + 11d
∴
37
= a + 11 (3)
∴
37
= a + 33
∴
37 – 33
= a
∴
a
= 4
Sn =
S12 =
=
= 246
∴
a = 4, S12 = 246
For given value of a12 Let’s use the formula
Lets find S12
Substitute n = 12, a = 4 & an i.e a12 = 37
Exercise 5.3 3(iii)
HOMEWORK
3) In an AP.
Sol:
i) Given a = 5, d = 3, an = 50, find n & Sn
For given AP:
a = 5,
d = 3,
an = 50
We know that,
an =
a + (n – 1) d
For given value of an,
Lets use the formula
∴ 50 =
5
+ (n – 1)
(3)
Substitute,
an = 50, a = 5 & d = 3
∴ 50 – 5 =
(n – 1) (3)
∴ 45 =
(n – 1) (3)
∴ 15 =
n – 1
∴ n =
16
Now lets find Sn
Now, Sn =
For Sn substitute,
a = 5, an = 50 & n = 16
8
∴ Sn =
∴ n = 16, Sn = 440
[5 + 50]
Exercise 5.3 3(i)
∴ S13 =
Sol:
ii) Given a = 7, a13 = 35, find d & S13
For given AP:
a = 7,
a13 = 35
We know that,
a13 =
a + 12d
For given value of a13,
Lets use the formula
∴ 35 =
7
+ 12d
∴ 35 – 7 =
12d
∴ 28 =
12d
∴ d =
∴ d =
7
3
Now lets find S13
Now, Sn =
For S13 substitute,
n = 13, a = 7 & a13 = 35
∴ S13
21
For a13 substitute,
a = 7 & a13 = 35
3) In an AP.
Exercise 5.3 3(ii)