Arithmetic
Progressions
∴ n = 12
4) How many terms of the AP: 9, 17, 25,… must be taken to
give a sum of 636 ?
Sol:
For given AP: 9, 17, 25, …
Means,
Sn = 636
Sn = 636
a = 9,
d = 17 – 9
= 8,
For given value of Sn,
Lets use the formula
We know that,
Sn =
Substitute,
a = 9, d = 8 & Sn = 636
∴ 636 =
∴ 636 × 2 =
∴ 1272 =
∴ 1272 =
8n2 + 10n – 1272
∴ 0 =
∴ 8n2 + 10n – 1272 = 0
Dividing throughout by 2, we get
∴ 4n2
+ 5n
– 636 = 0
It’s a quadratic equation, lets solve it by factorisation method
4 × 636
53
2 × 2 × 3 × 53
48
+
-
∴ 4n2
+ 53n – 48n
– 636 = 0
Take common from first two terms
∴ n
(4n + 53)
Take common from last two terms
– 12
(4n + 53)
= 0
∴ (4n + 53)
(n – 12)
= 0
or
n – 12 = 0
∴ 4n = – 53
n = 12
or
As ‘n’ cannot be negative
We need to find no. of terms i.e. value of ‘n’
n = 12
or
2×2
×
Exercise 5.3 4