Arithmetic

Progressions

• Sums based on a_n and S_n formula

∴ n = 7

(n – 7)

∴ n² – 5n – 14 = 0

14

7

2

–

+

– 7n + 2n

– 14 = 0

+ 2

(n – 7)

= 0

Sol:

vi) Given d = 2, a_n = 4, S_n = -14, find n & a.

For given AP:

d = 2,

a_n = 4,

S_n = -14

We know that,

a_n =

a + (n – 1) d

∴ 4 =

a

+ (n – 1)

(2)

∴ 4 =

a

+ 2n

– 2

∴ 4 + 2 =

a

+ 2n

∴ 6 =

a

+ 2n

∴ a =

6 – 2n

….(i)

S_n =

∴ -14 =

∴ -14 × 2 =

n

(10 – 2n)

∴ - 28 =

10n –

2n²

∴ 2n² – 10n – 28 = 0

Dividing throughout by 2, we get

∴ n²

∴ n

∴ (n – 7)

(n + 2)

= 0

• n – 7 = 0

or

n + 2 = 0

∴ n = 7

n = – 2

or

As n cannot be negative, n ≠ – 2

∴ a = 6 – 2(7)

∴ a = 6 – 14

∴ a = – 8

3) In an AP.

Exercise 5.3 3(viii)