SUM OF N TERMS OF SERIES

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Question 1 :

Find the general term and sum to n terms of the sequence

1, 4/3 , 7/9 , 10/27, . . . .

Solution :

1, 4/3 , 7/9 , 10/27, . . . .

a = 1, d = 3, r = 1/3

General term :

tn  =  ((a + (n โˆ’ 1)d)rnโˆ’1)

tn  =  ((1 + (n โˆ’ 1)3)(1/3)nโˆ’1)

  =  (1 + 3n - 3)(1/3)nโˆ’1

tn  =  (3n - 2)/3n โˆ’1

Sn  =  3n - (3n - 2)/2โ‹…3n-1 + (3n-1 - 1)/(8 โ‹… 3n-3)

Question 2 :

Find the value of n, if the sum to n terms of the series

โˆš3 + โˆš75 + โˆš243 + ยท ยท ยท is 435โˆš3.

Solution :

โˆš3 + โˆš75 + โˆš243 + ยท ยท ยท+ n terms

โˆš3 + 5โˆš3 + 9โˆš3 + ยท ยท ยท+ n terms

The given series is arithmetic series.

Sn  =  435โˆš3

(n/2)[2a+ (n-1)d]  = 435โˆš3

a = โˆš3, d = 4โˆš3

(n/2)[2โˆš3+ (n-1)(4โˆš3)]  = 435โˆš3

(n/2)[2โˆš3+ 4โˆš3n - 4โˆš3)]  = 435โˆš3

(n/2)[4โˆš3n - 2โˆš3)]  = 435โˆš3

n[2โˆš3n - โˆš3]  = 435โˆš3

Divide by โˆš3 on both sides

n[2n - 1]  = 435

2n2 - n - 435  =  0

(n - 15) (2n + 29)  =  0

n = 15, n = -29/2

Hence the value of n is 15.

Question 3 :

Show that the sum of (m + n)th and (m โˆ’ n)th term of an AP. is equal to twice the mth term

Solution :

tm+n  =  a + (m + n - 1)d  ------(1)

tm-n  =  a + (m - n - 1)d  ------(2)

(1) + (2)  

tm+n + tm-n  =  a + (m + n - 1)d + a + (m - n - 1)d 

  =  a + md + nd - d + a + md - nd - d 

  =  2a + 2md - 2d

  =  2 [a + (m-1) d]

  =  2 tm

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