F.5maths(Σ of Arithmetic Seq.)

52.
Firstly, consider a sequence 13, 14, 15, 16 ...... 50th term
The 50th term = 13 + 49*1 = 62
Then, 13 + 14 + 15 ......... + 62
= 50(13 + 62)/2
= 1875

Secondly, consider a sequence 15, 18, 21 ...... 60
No. of terms = [(60 +15)/3] + 1 = 16
Then, 15 + 18 + 21 + ...... + 60
= 16(15 + 60)/2
= 600

Hence, 13 + 14 -15 + 16 + 17 - 18 + ..... + 50th term
= (13 + 14 + 15 + ...... + 62) - 2(15 + 18 + 21 + ...... + 60)
= 1875 - 2(600)
= 675


53.
P : the first term = a, the common difference = d

Consider the sequence S(1), S(2), S(3) ......

S(n) = n[2a + (n - 1)d]/2
S(n + 1) = (n + 1)(2a + nd)/2

Let D be the common difference of the sequence S(1), S(2), S(3) ......
D = (n + 1)(2a + nd)/2 - n[2a + (n - 1)d]/2
D = (1/2)[(n + 1)(2a + nd) - n(2a + nd - d)]
D = (1/2)[2an + n²d + 2a + nd - 2an - n²d + nd]
D = (1/2)nd
d = 2D/n = constant

D is a constant, but n is NOT.
Hence, d is constant only when D = 0, i.e. d = 0


Common difference = (1/2)nd = constant
d = (2/n) * constant

Yes, I agree.

52 Divided the original sequence into 3 parts

13,16,19... 58,61 Sum = (17/2)(26 + 16 * 3) = 629

14,17,20,...,,59,62 Sum = (17/2)(28 + 16 * 3) = 646

-15,-18,-21....,-60 Sum = (16/2)(-30 + 15 * -3) = -600

The sum of the first 50 terms of the sequence = 675

53 Consider S(k + 1) and S(k)

S(k + 1) - S(k)

= T(k + 1)

= a + kd (where a is the first term and d is the common difference)

As Q forms the arithmetic sequence => a + kd = C (constant)

=> d = 0

The claim of the student is true.